solved examples of jntegration

- Antiderivatives
Example 1:  Find an antiderivative of f(x) = 2x.
Solution:  The function F(x) = x² is an antiderivative of f(x) since F ' (x) = 2x = f(x).
Example 2:  Find an antiderivative of f(x) = x⁴.
Solution:  F(x) = 1/5 x⁵ + C
Example 3:  Find an antiderivative of f(x) = 2x + x²
Solution:  F(x) = x² + 1/3 x³ + C
Example 4:  Find an antiderivative of f(x) = x e^x2
Solution:  F(x) = 1/2 e^x2 + C
Example 5:  Evaluate Integral [ x^2/3 ] dx.
Solution:  Integral [ x^2/3 ] dx = [ x^{2/3 + 1} / ( 2/3 + 1 ) ] + C = 3/5 x^5/3 + C
Example 6:  Evaluate Integral [ x^-3 ] dx
Solution:  Integral [ x^-3 ] dx = [ x^-2 / ( -2 ) ] + C = -1/2 x^-2 + C
Example 7:  Evaluate Integral [ 2 x² ] dx
Solution:  Integral [ 2 x² ] dx = 2 Integral [ x² ] dx = 2 [ x³ / 3 + C ] = 2/3 x³ + C
Example 8:  Evaluate Integral [ 2 x^-1/2 + 4x ] dx
Solution:  Integral [ 2x^-1/2 + 4x ] dx = Integral [ 2x^-1/2 ] dx + Integral [ 4x ] dx = 2 Integral [ x^-1/2 ] dx + 4 Integral [ x ] dx = 2 [ x^1/2 / ( 1/2 ) ] + 4 [ x² / 2 ] + C = 4x^1/2 + 2x² + C
Example 9:  Find the antiderivative F(x) of f(x) = 3x for which F(0) = 1
Solution:  F(x) = Integral [ 3x ] dx = 3 Integral [ x ] dx = 3 [ x² / 2 ] + C = 3/2 x² + C.   Now 1 = F(0) = 3/2 ( 0 )² + C = C, so F(x) = 3/2 x² + 1
Example 10:  The marginal cost C_M(x) of producing x kilograms of penicillin is C_M(x) = 2500 + 10x^3/2 dollars per kilogram.   If the start-up cost is $3500, what is the total cost function C(x)?
Solution:  Since C_M(x) = C ' (x), C(x) is an antiderivative of the marginal cost with C(0) = 3500.   Thus C(x) = Integral [ 2500 + 10x^3/2 ] dx = Integral [ 2500 ] dx + Integral [ 10x^3/2 ] dx = 2500 Integral [ 1 ] dx + 10 Integral [ x^3/2 ] dx = 2500 [ x ] + 10 [ 2/5 x^5/2 ] + C.    Now 3500 = C(0) = 2500 ( 0 ) + 4 ( 0 )^5/2 + C = C, so C(x) = 2500 x + 10 x^5/2 + 3500
Example 11:  Evaluate Integral [ 7 / x ] dx
Solution Integral [ 7 / x ] dx = 7 Integral [ 1 / x ] dx = 7 ln | x | + C
Example 12:  Evaluate Integral [ ( x + 1 ) / x ] dx
Solution:  Integral [ ( x + 1 ) / x ] dx = Integral [ x/x + 1/x ] dx = Integral [ 1 ] dx + Integral [ 1 / x ] dx = [ x ] + [ ln | x | ] + C
Example 13:  Evaluate Integral [ 3 e^2x ] dx
Solution:  Integral [ 3 e^2x ] dx = 3 Integral [ e^2x ] dx = 3 [ 1/2 e^2x ] + C = 3/2 e^2x + C
Example 14:  Evaluate Integral [ e^-3x ] dx
Solution:  Integral [ e^-3x ] dx = 1/(-3) e^-3x + C = -1/3 e^-3x + C
Example 15:  Find an antiderivative F(x) of f(x) = 2e^4x with F(0) = 8.
Solution:  F(x) = Integral [ 2 e^4x ] dx = 2 Integral [ e^4x ] dx = 2 [ 1/4 e^4x ] + C = 1/2 e^4x + C.   Then 8 = F(0) = 1/2 e^{4 ( 0 )} + C = 1/2 + C, so C = 15/2 and F(x) = 1/2 e^4x + 15/2

---

6.2, 6.3 - The Definite Integral and Area
Example 1:  Evaluate Integral₀¹ [ x ] dx
Solution:  Integral [ x ] dx = 1/2 x² + C, so Integral₀¹ [ x ] dx = [ 1/2 ( 1 )² + C ] - [ 1/2 ( 0 )² + C ] = 1/2
Example 2:  Evaluate Integral₁³ [ 2 / x ] dx
Solution:  Integral [ 2 / x ] dx = 2 Integral [ 1 / x ] dx = 2 [ ln | x | ] + C, so Integral₁³ [ 2 / x ] dx = [ 2 ln | 3 | + C ] - [ 2 ln | 1 | + C ] = 2 ln 3
Example 3:  Evaluate Integral₀¹ [ x² ] dx
Solution:  Integral [ x² ] dx = 1/3 x³ + C, so Integral₀¹ [ x² ] dx = [ 1/3 ( 1 )³ + C ] - [ 1/3 ( 0 )³ + C ] = 1/3
Example 4:  Evaluate Integral_-1¹ [ x³ ] dx
Solution:  Integral [ x³ ] dx = 1/4 x⁴ + C, so Integral_-1¹ [ x³ ] dx = [ 1/4 ( 1 )⁴ + C ] - [ 1/4 ( -1 )⁴ + C ] = 0
Example 5:  Evaluate Integral₀⁴ [ 2x ( x² + 1 )^1/2 ] dx
Solution:  Integral [ 2x ( x² + 1 )^1/2 ] dx = 2/3 ( x² + 1 )^3/2 + C (check this by taking the derivative!), so Integral₀⁴ [ 2x ( x² + 1 )^1/2 ] dx = [ 2/3 ( ( 4 )² + 1 )^3/2 + C ] - [ 2/3 ( ( 0 )² + 1 )^3/2 + C ] = 2/3 [ ( 17 )^3/2 - 1 ]
Example 6:  Find the area of the region bounded by the parabola y = x², the x-axis, and the vertical lines x = 0 and x = 1
Solution:  Area = Integral₀¹ [ x² ] dx = [ 1/3 ( 1 )³ + C ] - [ 1/3 ( 0 )³ + C ] = 1/3
Example 7:  Find the area of the region bounded by the graph of y = e^2x, the x-axis, and the vertical lines x = -2 and x = 1
Solution:  Area = Integral_-2¹ [ e^2x ] dx = [ 1/2 e^{2 ( 1 )} + C ] - [ 1/2 e^{2 ( -2 )} + C ] = 1/2 [ e² - e^-4 ] = 3.6854
Example 8:  Use a definite integral to find the area of a right triangle of height h and base b
Solution:  Put the right angle of the triangle at the origin, so the triangle is the region to the right of the y-axis, above the x-axis, and below the line connecting the points (0,h) and (b,0).   This last line has slope -h/b, so has equation (using slope-intercept form) y = -h/b x + h.   Then Integral [ -h/b x + h ] dx = -h/b ( 1/2 x² ) + h ( x ) + C.    Therefore Area = Integral₀^b [ -h/b x + h ] dx = [ -h/2b ( b )² + h ( b ) + C ] - [ -h/2b ( 0 )² + h ( 0 ) + C ] = -h/2 ( b ) + h ( b ) = 1/2 hb
Example 9:  Suppose the marginal cost of producing golf balls is given by
C_M(x) = 150 + 0.02 x dollars per gross If 500 gross are already produced, what is the cost of producing the next 500?
Solution:  We're asked to find C(1000) - C(500).   Since total cost is an antiderivative of marginal cost, this difference is equal to Integral₅₀₀¹⁰⁰⁰ [ C_M(x) ] dx = Integral₅₀₀¹⁰⁰⁰ [ 150 + 0.02 x ] dx = [ 150 ( 1000 ) + 0.02 ( 1/2 ( 1000 )² ) + C ] - [ 150 ( 500 ) + 0.02 ( 1/2 ( 500 )² ) + C ] = $82,500
Example 10:  Suppose marginal profit is P_M(x) = 200 - 9 x^1/2.   What is the profit earned from the sale of the 225th item to the 400th item?
Solution:  Profit is an antiderivative of marginal profit, so P(400) - P(225) = Integral₂₂₅⁴⁰⁰ [ 200 - 9 x^1/2 ] dx = [ 200 ( 400 ) - 9 ( 2/3 ( 400 )^3/2 ) + C ] - [ 200 ( 225 ) - 9 ( 2/3 ( 225 )^3/2 ) + C ] = $7,250
Example 11:  Between 1980 and 2000, the rate of oil consumption has risen 1% per year.   In 1980, the rate of consumption was 20 billion barrels per day.   How many barrels of oil were consumed in the 20 years between 1980 and 2000?
Solution:  We first need to find the function expressing the rate of consumption.   We're told that R ' (t) = 0.01 R(t), i.e., that the rate increased by 1% per year.   Thus the rate function is exponential of the form R(t) = R₀ e^{0.01 t}, and R₀ = 20, since the rate in 1980 was 20 billion barrels.   So the total oil consumed is given by Integral₀²⁰ [ R(t) ] dt = Integral₀²⁰ [ 20 e^{0.01 t} ] dt = 20 Integral₀²⁰ [ e^{0.01 t} ] dt = 20 { [ 1/0.01 e^{0.01 ( 20 )} + C ] - [ 1/0.01 e^{0.01 ( 0 )} + C ] } = 20 { 100 e^0.2 - 100 ] = 442.81 billion barrels

---

6.5 - Applications of the Definite Integral
Example 1:  Find the average value of the funtion y = x² between x = 1 and x = 3.
Solution:  [ Integral₁³ x² dx ] / ( 3 - 1 ) = 1/2 [ 1/3 x³ ]₁ ³ = 1/2 [ 1/3 ( 27 ) - 1/3 ( 1 ) ] = 26/6 = 13/3.
Example 2:  The daily output of a assembly line during the month of May is modelled by P(t) = 15 + 40t - t², where t is the t^th work day of the month of May.   What was the average productivity over the course of the 22 work days in May?
Solution:  [ Integral₀²² ( 15 + 40t - t² ) dt ] / ( 22 - 0 ) = 1/22 [ 15t + 20t² - 1/3 t³ ]₀²² = 293.67
Example 3:  The average annual per capita energy consumption (in millions of BTUs) has grown 2% per year since 1940.    In 1940 the average annual per capita energy consumption was 181 million BTUs.   What was the average annual per capita energy consumption between the years 1940 and 1970?
Solution:  Let B(t) = the average annual per capita energy consumption t years after 1940.   Then we want to know the average value of B(t) between t = 0 and t = 30.   First, we need to know what B(t) is.   Since it is growing as a percentage of its present value, i.e., B ' (t) = 0.02 B(t), B(t) is exponential.   Therefore B(t) = B₀ e^{0.02 t} = 181 e^{0.02 t}.   Hence the average value is [ Integral₀³⁰ 181 e^{0.02 t} dt ] / [ 30 - 0 ] = 1/30 [ 181/0.02 e^{0.02 t} ]₀³⁰ = 248 million BTUs.
Example 4:  Find the consumer's surplus for the demand function D(x) = ( x - 5 )² when q = 3.
Solution:  When q = 3, p = D(3) = ( -2 )² = 4, so the consumer's surplus is [ Integral₀³ ( x² - 10x + 25 ) dx ] - [ ( 3 )( 4 ) ] = [ 1/3 x³ - 5 x² + 25 x ]₀³ - 12 = $27.
Example 5:  Find the producer's surplus for the supply function S(x) = x² + x + 3 when q = 3.
Solution:  When q = 3, p = S(3) = 15, so the producer's surplus is [ ( 3 )( 15 ) ] - [ Integral₀³ ( x² + x + 3 ) dx ] = 45 - [ 1/3 x³ + 1/2 x² + 3 x ]₀³ = $22.50.
Example 6:  Suppose the demand function is D(x) = ( x - 5 )² and the supply function is S(x) = x² + x + 3 for a certain item.
a. What is the equilibrium point?
b. What is the consumer's surplus at this point?
c. What is the producer's surplus at this point?
Solution:  a. Set S(x) = D(x).   So x² + x + 3 = x² - 10 x + 25, i.e., 11 x = 22.   Thus q_E = 22/11 = 2 and p_E = S(2) = $9.   Therefore the equilibrium point is (2,$9).
b. Consumer's surplus = [ Integral₀² ( x² - 10x + 25 ) dx ] - [ ( 2 )( 9 ) ] = [ 1/3 x³ - 5 x² + 25 x ]₀² - [ 18 ] = $14.67.
c. Producer's surplus = [ ( 2 )( 9 ) ] - [ Integral₀ ² ( x² + x + 3 ) dx ] = [ 18 ] - [ 1/3 x³ + 1/2 x² + 3 x ]₀² = $7.33.
Example 7:  Suppose money is flowing continuously into a savings account at an annual rate of $1000 per year at an interest rate of 8% compounded continuously.
a. How much money is in the account after 5 years?
b. How much money is in the account after 15 years?
Solution:  a. Integral₀⁵ [ 1000 e^{0.08 t} ] dt = [ 1000/0.08 e^{0.08 t} ]₀ ⁵ = $6147.81.
b. Integral₀¹⁵ [ 1000 e^{0.08 t} ] dt = [ 1000/0.08 e^{0.08 t} ]₀ ¹⁵ = $29,001.46.
Example 8:  Suppose P₀ dollars is invested each year into a savings account paying 8% interest compounded continuously over a period of 20 years.   If we want to have $10,000 in the account at the end of the 20 years, what should P₀ be to ensure this?
Solution:  We want 10000 = Integral₀²⁰ [ P₀ e^{0.08 t} ] dt = P₀ [ 1/0.08 e^{0.08 t} ]₀²⁰ = 49.4 P₀, so we need P₀ = 10000 / 49.4 = $202.38.

---

9.1 - Substitution
Example 1:  If u = x³, find its differential du.
Solution:  du = [ 3 x² ] dx
Example 2:  If u = ln x, find its differential du.
Solution:  du = [ 1 / x ] dx
Example 3:  Evaluate Integral [ 2x e^x2 ] dx
Solution:  Let u = x², so du = 2x dx.  Then Integral [ 2x e^x2 ] dx = Integral e^u du = e^u + C = e^x2 + C.
Example 4:  Evaluate Integral [ 2x / ( 1 + x² ) ] dx
Solution:  Let u = 1 + x², so du = [ 2x ] dx.   Then Integral [ 2x / ( 1 + x² ) ] dx = Integral [ 1 / u ] du = ln | u | + C = ln | 1 + x² | + C
Example 5:  Evaluate Integral [ 3 x² / ( 1 + x³ )² ] dx
Solution:  Let u = 1 + x³, so du = [ 3 x² ] dx.  Then Integral [ 3 x² / ( 1 + x³ )² ] dx = Integral [ 1 / u² ] du = Integral [ u^-2 ] du = u^-1 / ( -1 ) + C = -( 1 + x³ )^-1 + C
Example 6:  Evaluate Integral [ ln( 3x ) / x ] dx.
Solution:  Let u = ln( 3x ), so du = [ 1 / ( 3x ) ]( 3 ) dx = [ 1 / x ] dx.  Then Integral [ ln( 3x ) / x ] dx = Integral [ u ] du = u² / 2 + C = 1/2 ( ln( 3x ) )² + C
Example 7:  Evaluate Integral [ x e^x2 ] dx
Solution:  Let u = x², so du = 2x dx and x dx = 1/2 du.  Then Integral [ x e^x2 ] dx = Integral [ e^u ] ( 1/2 du ) = 1/2 Integral e^u du = 1/2 e^u + C = 1/2 e^x2 + C
Example 8:  Evaluate Integral [ e^x / ( 4 + e^x ) ] dx
Solution:  Let u = 4 + e^x, so du = e^x dx.  Then Integral [ e^x / ( 4 + e^x ) ] dx = Integral [ 1 / u ] du = ln | u | + C = ln | 4 + e^x | + C
Example 9:  Evaluate Integral [ 1 / ( x + 3 ) ] dx
Solution:  Let u = x + 3, so du = dx and Integral [ 1 / ( x + 3 ) ] dx = Integral [ 1 / u ] du = ln | u | + C = ln | x + 3 | + C
Example 10:  Evaluate Integral [ x² ( x³ + 1 )¹⁰ ] dx
Solution:  Let u = x³ + 1, so du = 3x² dx and x² dx = 1/3 du.  Thus Integral [ x² ( x³ + 1 )¹⁰ ] dx = Integral [ u¹⁰ ] ( 1/3 du ) = 1/3 [ u¹¹ / 11 ] + C = 1/33 ( x³ + 1 )¹¹ + C
Example 11:  Evaluate Integral [ x⁴ e^x5 ] dx
Solution:  Let u = x⁵, so du = 5x⁴ dx and x⁴ dx = 1/5 du.  Therefore Integral [ x⁴ e^x5 ] dx = Integral [ e^u ] ( 1/5 du ) = 1/5 [ e^u ] + C = 1/5 e^x5 + C
Example 12:  Evaluate Integral [ 1 / ( x ln( x² ) ) ] dx
Solution:  Let u = ln( x² ), so du = [ 1 / x² ] ( 2x ) dx = [ 2 / x ] dx and 1 / x dx = 1/2 du.  Then Integral [ 1 / ( x ln( x² ) ) ] dx = Integral [ 1 / u ] ( 1/2 du ) = 1/2 [ ln | u | ] + C = 1/2 ln | ln( x² ) | + C
Example 13:  Evaluate Integral [ x ( 4x² + 9 )^1/2 ] dx
Solution:  Let u = 4x² + 9, so du = 8x dx and x dx = 1/8 du.  Then Integral [ x ( 4x² + 9 )^1/2 ] dx = Integral [ u^1/2 ] ( 1/8 du ) = 1/8 [ u^3/2 / ( 3/2 ) ] + C = 1/12 ( 4x² + 9 )^3/2 + C
Example 14:  Evaluate Integral [ ( 7x + 1 )^1/2 ] dx
Solution:  Let u = 7x + 1, so du = 7 dx and dx = 1/7 du.  Then Integral [ ( 7x + 1 )^1/2 ] dx = Integral [ u^1/2 ] ( 1/7 du ) = 1/7 [ u^3/2 / ( 3/2 ) ] + C = 2/21 ( 7x + 1 )^3/2 + C
Example 15:  Evaluate Integral [ x ( x + 1 )^1/2 ] dx
Solution:  Let u = x + 1, so x = u - 1 and dx = du.  Then Integral [ x ( x + 1 )^1/2 ] dx = Integral [ ( u - 1 ) u^1/2 ] du = Integral [ u^3/2 - u^1/2 ] du = u^5/2 / ( 5/2 ) - u^3/2 / ( 3/2 ) + C = 2/5 ( x + 1 )^5/2 - 2/3 ( x + 1 )^3/2 + C

---

9.2 - Integration by Parts
Example 1:  Compute Integral [ x e^x ] dx.
Solution A:  Clearly this is not of the form for any of the standard integrals we have memorized.   Our next idea would be to do a substitution, but letting u = x doesn't do enough and u = e^x doesn't address the problem of the extra x we have in the integrand.    So this problem requires integration by parts, which means we need to decide what to make our f(x) and what should be g(x) in the expression x e^x.   Following hint 1, we select g(x) first and so try g(x) = e^x, in which case f(x) = x.   Then G(x) = e^x and f ' (x) = 1, so that
Integral [ x e^x ] dx = ( x ) ( e^x ) - Integral [ ( 1 ) ( e^x ) ] dx = x e^x - [ e^x + C ] = x e^x - e^x + C Solution B:  Suppose in this example that we had chosen f(x) and g(x) differently as f(x) = e^x and g(x) = x.   Then f ' (x) = e^x while G(x) = x² / 2, so that G(x) is "more complicated" than g(x) was which is our first indication that this might not work.   But if we continue we would have
Integral [ x e^x ] dx = ( e^x ) ( x² / 2 ) - Integral [ ( e^x ) ( x² / 2 ) ] dx, which now leaves us to compute Integral [ 1/2 x² e^x ] dx. However, this isn't a straight-forward integral and is in fact more complicated and "ugly" than what we started with, so with this second clue we would think to try different choices for f(x) and g(x) in our original problem, and thus obtain our solution in part A above. Example 2:  Evaluate Integral [ x ln x ] dx.
Solution:  Let us examine several choices, as follows.
Choice A:  Let f(x) = 1 and g(x) = x ln x.   This will not work because we are back to our original integral, in which we do not know how to integrate g(x) = x ln x.
Choice B:  We let f(x) = x ln x and g(x) = 1.   Then f ' (x) = ( 1 ) ln x + x ( 1 / x ) = ln x + 1 and G(x) = x.   Using the integration by parts formula, we have
Integral [ x ln x ] dx = ( x ln x ) ( x ) - Integral [ ( ln x + 1 ) ( x ) ] dx = x² ln x - Integral [ x ln x + x ] dx = x² ln x - [ Integral [ x ln x ] dx + x² / 2 + C ]
Now we observe that Integral [ x ln x ] dx appears on both sides of the equation, but occurs with a negative sign on the right.   So if we add Integral [ x ln x ] dx to both sides, we have 2 Integral [ x ln x ] dx = x² ln x - 1/2 x² + C
so that (dividing both sides by 2)
Integral [ x ln x ] dx = 1/2 x² ln x - 1/4 x² + C.
This method worked, but a third option below is even easier.   Remember that in math, there is only one right answer, but there might be several ways of getting there! Choice C:  We let f(x) = ln x and g(x) = x.   Then f ' (x) = 1 / x and G(x) = x² / 2, so the integration by parts formula says
Integral [ x ln x ] dx = ( ln x ) ( x² / 2 ) - Integral [ ( 1 / x ) ( x² / 2 ) ] dx = 1/2 x² ln x - Integral [ 1/2 x ] dx = 1/2 x² ln x - 1/2 [ x² / 2 + C ] = 1/2 x² ln x - 1/4 x² + C. Example 3:  Evaluate Integral [ x ( x + 1 )^1/2 ] dx.
Solution:  We let f(x) = x and g(x) = ( x + 1 )^1/2, so f ' (x) = 1 and G(x) = 2/3 ( x + 1 )^3/2.   Then
Integral [ x ( x + 1 )^1/2 ] dx = ( x ) [ 2/3 ( x + 1 )^3/2 ] - Integral [ ( 1 ) ( 2/3 [ x + 1 ]^3/2 ) ] dx = 2/3 x ( x + 1 )^3/2 - 2/3 [ 2/5 ( x + 1 )^5/2 + C ]
= 2/3 x ( x + 1 )^3/2 - 4/15 ( x + 1 )^5/2 + C. Example 4:  Evaluate Integral₁² [ ln x ] dx.
Solution:  First we need to compute Integral [ ln x ] dx.    Let f(x) = ln x and g(x) = 1 so that f ' (x) = 1 / x and G(x) = x.    Then Integral [ ln x ] dx = ( ln x ) ( x ) - Integral [ ( 1 / x ) ( x ) ] dx = x ln x - Integral [ 1 ] dx = x ln x - x + C.   Thus Integral₁² [ ln x ] dx = [ x ln x - x ]₁ ² = [ 2 ln 2 - 2 ] - [ 1 ln 1 - 1 ] = 2 ln 2 - 2 - [ - 1 ] = 2 ln 2 - 1.
Example 5:  Evaluate Integral [ x² e^x ] dx
Solution:  Let f(x) = x² and g(x) = e^x so f ' (x) = 2x and G(x) = e^x.   Then
Integral [ x² e^x ] dx = ( x² ) ( e^x ) - Integral [ ( 2x ) ( e^x ) ] dx = x² e^x - 2 Integral [ x e^x ] dx.
Now we can either recall from Example 1 what Integral [ x e^x ] dx is or we may use integration by parts a second time, with f(x) = x and g(x) = e^x (so f ' (x) = 1 and G(x) = e^x) to obtain Integral [ x² e^x ] dx = x² e^x - 2 [ x e^x - e^x + C ] = x² e^x - 2x e^x + 2 e^x + C.

---

9.6, 12.2 - Improper Integrals and Probability
Example 1:  Find the area of the region under the graph of y = 1 / x² over the interval [1,infinity).
Solution:  Integral₁^infinity [ x^-2 ] dx = lim_{b -> infinity} Integral₁ ^b [ x^-2 ] dx = lim_{b -> infinity} [ -x^-1 ]₁^b = lim_{b -> infinity} [ ( -1 / b ) + ( 1 ) ] = 1
Example 2:  Find the area of the region under the graph of y = 1 / x over the interval [1,infinity).
Solution:  Integral₁^infinity [ x^-1 ] dx = lim_{b -> infinity} Integral₁ ^b [ x^-1 ] dx = lim_{b -> infinity} [ ln | x | ]₁^b = lim_{b -> infinity} [ ( ln b ) - ( ln 1 ) ] does not exist, since ln x -> infinity as x -> infinity.
Example 3:  Evaluate Integral₀ ^infinity [ 2 e^{-2 x} ] dx
Solution:  Integral₀^infinity [ 2 e^{-2 x} ] dx = lim_{b -> infinity} Integral₀ ^b [ 2 e^{-2 x} ] dx = lim_{b -> infinity} [ -e^{-2 x} ]₀^b = lim_{b -> infinity} [ ( -e^-2b ) - ( -e⁰ ) ] = 0 + 1 = 1.
Example 4:  Evaluate Integral_-infinity ^infinity [ x e^-x2 ] dx
Solution:  Integral_-infinity^infinity [ x e^-x2 ] dx = Integral_-infinity ⁰ [ x e^-x2 ] dx + Integral₀ ^infinity [ x e^-x2 ] dx = lim_{a -> -infinity} Integral_a⁰ [ x e^-x2 ] dx + lim_{b -> infinity} Integral₀^b [ x e^-x2 ] dx = lim_{a -> -infinity} [ -1/2 e^-x2 ]_a⁰ + lim_{b -> infinity} [ -1/2 e^-x2 ]₀^b = lim_{a -> -infinity} [ ( -1/2 ) - ( -1/2 e^-a2 ) ] + lim_{b -> infinity} [ ( -1/2 e^-b2 ) - ( -1/2 ) ] = [ -1/2 + 0 ] + [ 0 + 1/2 ] = 0.
Example 5:  What is the probability of drawing an ace from a well-shuffled deck of playing cards?
Solution:  There are a total of 4 aces out of 52 cards, so the probability is 4/52 = 1/13 = 0.077.
Example 6:  Suppose that x is the arrival time of a bus at a bus stop between 2pm and 5pm.  What is the probability that the bus arrives between 4pm and 5pm?
Solution:  Suppose the bus is as likely to arrive at one moment as it is to arrive at another, so x is uniformly distributed and f(x) = 1 / ( 5 - 2 ) = 1/3.   Then P( 4 < x < 5 ) = Integral₄⁵ [ 1/3 ] dx = [ 1/3 x ]₄⁵ = 5/3 - 4/3 = 1/3.
Example 7:  Verify that f(x) = 3/117 x² is a probability density function for x on the interval [2,5].
Solution:  Clearly f(x) > 0 on the interval, so we only need to show that 1 = Integral₂⁵ [ 3/117 x² ] dx = 3/117 [ 1/3 x³ ]₂⁵ = 1/117 [ 125 - 8 ] = 1, as desired.
Example 8:  A company produces transistors.  It determines that the life-span t of a transistor is between 3 and 6 years with a probability density function f(t) = 24 t^-3.
a.  Find P( t < 4 )
b.  Find P( 4 < t < 5 )
Solution:  a. P( t < 4 ) = P( 3 < t < 4 ) = Integral₃⁴ [ 24 t^-3 ] dt = 24 [ t^-2 / ( -2 ) ]₃⁴ = 0.58
b. P( 4 < t < 5 ) = Integral₄⁵ [ 24 t^-3 ] dt = 24 [ t^-2 / ( -2 ) ]₄ ⁵ = 0.27
Example 9:  Find k so that f(x) = k x² is a probability density function for x on [2,5].
Solution:  We need 1 = Integral₂⁵ [ k x² ] dx = k [ x³ / 3 ]₂⁵ = k [ 125/3 - 8/3 ] = 117/3 k.   Thus k = 3/117.
Example 10:  The distance x between successive cars on a highway has probability density function f(x) = k e^{-k x}, for 0 < x < infinity, where k = 1/a and a = average distance between successive cars.   If a = 166 feet, what is the probability that x is less than 50 feet?
Solution:  P( x < 50 ) = Integral₀⁵⁰ [ 1/166 e^{-1/166 x} ] dx = [ -e^{-1/166 x} ]₀⁵⁰ = 0.26
Example 11:  Given the probability density function f(x) = 1/2 x for x on the interval [0,2], find the mean (expected value) of x.
Solution:  m = Integral₀ ² [ x ^. ( 1/2 x ) ] dx = 1/2 Integral₀ ² [ x² ] dx = 1/2 [ x³ / 3 ]₀² = 1/2 [ 8/3 - 0 ] = 8/6 = 4/3

---

7.1, 7.2 - Functions of Several Variables and Partial Derivatives
Example 0:  Functions of several variables are used often in our everyday lives. For instance, in the winter we are very interested in the wind chill factor, which uses both the actual, raw temperature outside and the current wind speed to tell us how cold it actually feels. In the summer, the heat index combines the temperature and relative humidity to measure how hot it really feels outside.
Example 1:  A company makes two items, guns and butter.   It sells each gun for $4 and each pound of butter for $6.  Find the company's total revenue function and evaluate it when it sells 25 guns and 10 pounds of butter.
Solution:  R(x,y) = 4x + 6y, where x = # guns and y = pounds of butter.  So R(25,10) = 4(25) + 6(10) = 160.
Example 2:  The total cost of a company, in thousands of dollars, is given by C(x,y,z,w) = 4x² + 5y + z - ln( w + 1 ), where x = money spent on labor, y = money spent on raw materials, z = money spent on advertising, and w = money spent on machinery.  Find C(3,2,0,10).
Solution:  C(3,2,0,10) = 4(3)² + 5(2) + (0) - ln( (10) + 1 ) = 36 + 10 - ln 11 = 43.6 thousands of dollars.
Example 3:  The wind chill at temperature T and wind speed v is given by W(T,v) = 91.4 - [ ( 10.45 + 6.68 v^1/2 - 0.477 v )( 457 - 5 T ) ] / 110, where T is the current temperature in degrees Fahrenheit and v is the wind speed in miles per hour.
a.  Find the wind chill when T = 30^o F, v = 20 mph.
b.  Find the wind chill when T = 20^o F, v = 20 mph.
c.  Find the wind chill when T = 20^o F, v = 40 mph.
Solution:  a. W(30,20) = 5.49
b. W(20,20) = -8.51
c. W(20,40) = -17.71
Example 4:  Find the first order partial derivatives of f(x,y) = x²y³ + xy + 4y².
Solution:  df/dx = (2x)y³ + (1)y + 0 = 2xy³ + y and
df/dy = x²(3y²) + x(1) + 8y = 3x²y² + x + 8y.
Example 5:  Find the first order partial derivatives of f(x,y,z) = x² - xy + y² + 2yz + 2z² + z.
Solution:  df/dx = 2x - y
df/dy = -x + 2y + 2z
df/dz = 2y + 4z + 1
Example 6:  Find the first order partial derivatives of f(x,y) = 3x²y + xy.
Solution:  df/dx = 3(2x)y + (1)y = 6xy + y
df/dy = 3x²(1) + x(1) = 3x² + x
Example 7:  Find the first order partial derivatives of f(x,y) = e^xy + y ln x
Solution:  df/dx = e^xy[(1)y] + y( 1/x ) = y e^xy + y/x
df/dy = e^xy[x(1)] + (1)ln x = x e^xy + ln x
Example 8:  Find the marginal productivity of labor for the Cobb-Douglas production function p(x,y) = 50 x^2/3 y^1/3, where x = # units of labor and y = # units of capital, when x = 125 and y = 64.
Solution:  The marginal productivity of labor is dp/dx = 50(2/3 x^-1/3)y^1/3, so dp/dx (125,64) = 100/3 (125)^-1/3 (64)^1/3 = 80/3
Example 9:  Find d² f / dy dx, where f(x,y) = 3xy² + 2xy + x².
Solution:  df/dx = 3(1)y² + 2(1)y + 2x = 3y² + 2y + 2x, so d² f / dy dx = d/dy [ df/dx ] = d/dy [ 3y² + 2y + 2x ] = 6y + 2
Example 10:  Find all second order partial derivatives of f(x,y) = x² y³ + x⁴ y + x e^y.
Solution:  df/dx = (2x)y³ + (4x³)y + (1)e^y and df/dy = x²(3y²) + x⁴(1) + x e^y.  Thus
d²f/dx² = d/dx [ 2xy³ + 4x³y + e^y ] = 2(1)y³ + 4(3x²)y + 0 = 2y³ + 12x²y
d²f/dydx = d/dy [ 2xy³ + 4x³y + e^y ] = 2x(3y²) + 4x³(1) + e^{y = 6xy2 + 4x3 + ey}
d²f/dxdy = d/dx [ 3x²y² + x⁴ + x e^y ] = 3(2x)y² + 4x³ + (1)e^y = 6xy² + 4x³ + e^y = d²f/dydx
d²f/dy² = d/dy [ 3x²y² + x⁴ + x e^y ] = 3x²(2y) + 0 + x(e^y) = 6x²y + x e^y

---

7.3 - Maxima and Minima of Functions of Two Variables
Example 1:  Find all first and second order partial derivatives of f(x,y) = 5x²y - 7xy + 2y.
Solution:  df/dx = 5(2x)y - 7(1)y + 0 = 10xy - 7y
df/dy = 5x²(1) - 7x(1) + 2(1) = 5x² - 7x + 2
d²f/dx² = 10(1)y - 0 = 10y
d²f/dydx = 10x(1) - 7(1) = 10x - 7
d²f/dxdy = 5(2x) - 7(1) + 0 = 10x - 7
d²f/dy² = 0 - 0 + 0 = 0
Example 2:  Use the First Derivative Test to locate the relative minimum of f(x,y) = 3x² - 4xy + 3y² + 8x - 17y + 30.
Solution:  df/dx = 6x - 4y + 8 = 0 when 4y = 6x + 8 or y = 3/2 x + 2.
df/dy = -4x + 6y - 17 = -4x + 6(3/2 x + 2) - 17 = -4x + 9x + 12 - 17 = 5x - 5 = 0 when x = 1.
When x = 1, we also have y = 3/2 + 2 = 7/2.  So f has its relative minimum at the point (1,7/2).
Example 3:  Use the First Derivative Test to locate the relative minimum of f(x,y) = 1/2 x² + y² - 3x + 2y - 5.
Solution:  df/dx = x - 3 = 0 when x = 3
df/dy = 2y + 2 = 0 when y = -1
So f has its relative minimum at the point (3,-1).
Example 4:  Use the First and Second Derivative Tests to find the relative maxima and/or minima of f(x,y) = y³ - x² + 6x - 12y + 5.
Solution:  df/dx = -2x + 6 = 0 when x = 3
df/dy = 3y² - 12 = 3(y² - 4) = 3(y - 2)(y + 2) = 0 when y = +2
So the possible extreme points are at (3,2) and (3,-2).
Now d²f/dx² = -2
d²f/dydx = 0
d²f/dxdy = 0
d²f/dy² = 6y
So D(x,y) = (d²f/dx²)(d²f/dy²) - (d²f/dydx)²
At (3,2), we have D(3,2) = (-2)[6(2)] = -24 < 0, so NEITHER
At (3,-2), we have D(3,-2) = (-2)[6(-2)] = +24 > 0 and d²f/dx² = -2 < 0, so RELATIVE MAXIMUM
Example 5:  Use the First Derivative Test to locate the relative minimum of f(x,y) = -x² - 8xy - y².
Solution:  df/dx = -2x - 8y = 0 when x = -4y
df/dy = -8x - 2y = -8(-4y) - 2y = 30y = 0 when y = 0.
So the only possible extreme point is when x = -4(0) = 0 and y = 0.
Now d²f/dx² = -2
d²f/dydx = -8
d²f/dxdy = -8
d²f/dy² = -2
So D(x,y) = (-2)(-2) - (-8)² = 4 - 64 = -60 < 0
so f has NEITHER a relative max or min at (0,0).
Example 6:  A firm makes two kinds of golf balls.  One sells for $3 each and the other sells for $2 each.  Thus the revenue from selling x (thousand) balls at $3 and y (thousand) balls at $2 is
R(x,y) = 3x + 2y The company has determined that the total cost function of producing x thousand balls of the first type and y thousand balls of the second is given by
C(x,y) = 2x² - 2xy + y² - 9x + 6y + 7 Find the number of balls of each type that should be made in order to maximize this company's profits. Solution:  P(x,y) = R(x,y) - C(x,y) = [ 3x + 2y ] - [ 2x² - 2xy + y² - 9x + 6y + 7 ] = -2x² + 2xy - y² + 12x - 4y - 7.  Now
dP/dx = -4x + 2y + 12 = 0 when 2y = 4x - 12
dP/dy = 2x - 2y - 4 = 0 when 2y = 2x - 4
so both are zero when 4x - 12 = 2x - 4, i.e., when 2x = 8, so x = 4 and y = 1/2 [ 2(4) - 4 ] = 2
Thus our possible max occurs at the point (4,2).
Now d²P/dx² = -4, d²P/dydx = 2, and d²P/dy² = -2, so
D(x,y) = ( -4 )( -2 ) - ( 2 )² = 8 - 4 = 4.
Thus D(4,2) = 4 > 0 and d²P/dx² (4,2) = -4 < 0
so profit is maximized when x = 4 (thousand) and y = 2 (thousand).

---

7.4 - Lagrange Multipliers and Constrained Optimization
Lecture 1
Example 0:  The autumn demand for home insulation depends on the unit cost x for heating fuel and the unit cost y for insulation.  Suppose the demand for insulation is given by
d(x,y) = -4x² + 2xy - 6y² + 28x + 62y + 850 For what values of x and y is the demand maximum?
Solution:  dd/dx = -8x + 2y + 28 = 0 when y = 4x - 14
dd/dy = 2x - 12y + 62 = 2x - 12( 4x - 14 ) + 62 = -46x + 230 = 0
when x = 5 and y = 4( 5 ) - 14 = 6
Now d²d/dx² = -8, d²d/dydx = 2, and d²d/dy² = -12
So D(x,y) = ( -8 )( -12 ) - ( 2 )² = 92
Since D(5,6) = 92 > 0 and d²d/dx² (5,6) = -8 < 0
demand is maximum when x = 5 and y = 6.
Example 1:  Maximize f(x,y) = 2xy subject to the constraint g(x,y) = 2x + 3y - 6 = 0.
Solution:  Set F(x,y,l) = f(x,y) - l g(x,y) = [ 2xy ] - l [ 2x + 3y - 6 ]
dF/dx = 2y - 2l = 0 when l = y
dF/dy = 2x - 3l = 0 when l = 2/3 x
So y = l = 2/3 x, which we plug into
dF/dl = -2x - 3y + 6 = -2x -3( 2/3 x ) + 6 = -4x + 6 = 0
when x = 3/2, so y = 2/3 ( 3/2 ) = 1.
So f is maximize at ( 3/2 , 1 ), with value f(3/2,1) = 3
Example 2:  Minimize f(x,y) = x² + y² subject to g(x,y) = xy - 2 = 0.
Solution:  Set F(x,y,l) = [ x² + y² ] - l [ xy - 2 ]
dF/dx = 2x - yl = 0 when l = 2x/y
dF/dy = 2y - xl = 0 when l = 2y/x
so 2x/y = 2y/x, i.e., x² = y² (by cross multiplying)
so y = +x, which we now plug into
dF/dl = -xy + 2 = -x² + 2 = 0 when y = x
or x² + 2 = 0 when y = -x, but this can't happen so y = x.
So y = x = +2^1/2, making the possible minimimum points at either ( 2^1/2,2^1/2 ) or ( -2^1/2,-2^1/2 )
Now f(+2^1/2,+2^1/2) = 4 in both cases, so both are minima.
Example 3:  Maximize P(r,s) = r - s subject to r² + s² = 4.
Solution:  Set F(r,s,l) = [ r - s ] - l [ r² + s² - 4 ]
dF/dr = 1 - 2rl = 0 when l = 1/2r
dF/ds = -1 - 2sl = 0 when l = -1/2s
so 2s = -2r, i.e., s = -r, which we now use in
dF/dl = -r² - s² + 4 = -r² - ( -r )² + 4 = -2r² + 4 = 0
when r = +2^1/2 and s = +2^1/2
Now we want to maximize, so consider
P( 2^1/2,-2^1/2 ) = 2^3/2
P( -2^1/2,2^1/2 ) = -2^3/2
so P is maximized at ( 2^1/2,-2^1/2 )
Example 4:  Minimize and maximize f(x,y) = 2x² + 3y² subject to x⁴ + y² = 1.
Solution:  Set F(x,y,l) = [ 2x² + 3y² ] - l [ x⁴ + y² - 1 ]
dF/dx = 4x - 4x³ l = 4x ( 1 - x² l ) = 0 when x = 0 or l = 1/x²
dF/dy = 6y - 2y l = 2y ( 3 - l ) = 0 when y = 0 or l = 3
When x = 0, 0⁴ + y² = 1 implies y = +1
when y = 0, x⁴ + 0² = 1 implies x = +1
otherwise 3 = l = 1/x² so x = +( 1/3 )^1/2 and y = +( 8/9 )^1/2
Now f(0,+1) = 3
f(+1,0) = 2
f(+( 1/3 )^1/2,+( 8/9 )^1/2) = 2/3 + 8/3 = 10/3
so the max is at (+( 1/3 )^1/2,+( 8/9 )^1/2) and the min is at (+1,0).
Example 5:  Minimize x² + 2xy + 2y² + x + y subject to 1 - x - 2y = 0.
Solution:  Set F(x,y,l) = [ x² + 2xy + 2y² + x + y ] - l [ 1 - x - 2y ]
dF/dx = 2x + 2y + 1 + l = 0 when l = -1 - 2y - 2x
dF/dy = 2x + 4y + 1 + 2l = 0 when l = -1/2 - x - 2y
so -1 - 2y - 2x = -1/2 - x - 2y, so -x = 1/2 or x = -1/2
dF/dl = -1 + x + 2y = -1 + ( -1/2 ) + 2y = -3/2 + 2y = 0 when y = 3/4
so the minimum occurs when x = -1/2 and y = 3/4
Lecture 2
Example 1:  Maximize f(x,y) = 4 - x² - y² subject to x + 2y = 10.
Solution:  Set F(x,y,l) = [ 4 - x² - y² ] - l [ x + 2y - 10 ]
dF/dx = -2x - l = 0 when l = -2x
dF/dy = -2y - 2l = 0 when l = -y, so y = 2x
dF/dl = -x - 2y + 10 = -5x + 10 = 0 when x = 2 and y = 4
Example 2:  Suppose the production function of a company is given by f(x,y) = 8 x^1/4 y^3/4, where x is the number of units of labor and y is the number of units of capital.  If x + y = 64, what is the maximum productivity?
Solution:  Set F(x,y,l) = [ 8 x^1/4 y^3/4 ] - l [ x + y - 64 ]
dF/dx = 2 x^-3/4 y^3/4 - l = 0 when l = 2 x^-3/4 y^3/4
dF/dy = 6 x^1/4 y^-1/4 - l = 0 when l = 6 x^1/4 y^-1/4
so 2 x^-3/4 y^3/4 = 6 x^1/4 y^-1/4, i.e., y = 3x
dF/dl = - x - y + 64 = - 4x + 64 = 0 when x = 16 and y = 48
then the productivity is f(16,48) = 291.78
Example 3:  The production function of (another) company is f(x,y) = 4 x^1/2 y^1/2, where x is the number of units of labor and y the number of units of capital.  If labor costs $300 per unit and capital costs $100 per unit and the firm can only invest $18,000, what values of x and y maximize productivity?
Solution:  Set F(x,y,l) = [ 4 x^1/2 y^1/2 ] - l [ 300x + 100y - 18000 ]
dF/dx = 2 x^-1/2 y^1/2 - 300l = 0 when l = 1/150 x^-1/2 y^1/2
dF/dy = 2 x^1/2 y^-1/2 - 100l = 0 when l = 1/50 x^1/2 y^-1/2
so 1/150 x^-1/2 y^1/2 = 1/50 x^1/2 y^-1/2, i.e., y = 3x
dF/dl = -300x - 100y + 18000 = -600x + 18000 = 0 when x = 30 and y = 90.
Example 4:  The total sales S of a one-product firm are given by S(L,M) = 2ML - L², where M is the cost of materials and L is the cost of labor.  Find the maximum sales subject to the budget constraint M + L = 60.
Solution:  Set F(L,M,l) = [ 2ML - L² ] - l [ M + L - 60 ]
dF/dL = 2M - 2L - l = 0 when l = 2M - 2L
dF/dM = 2L - l = 0 when l = 2L
so 2L = 2M - 2L, i.e., M = 2L
dF/dl = -M - L + 60 = -3L + 60 = 0 when L = 20 and M = 40.
Example 5:  A product can be made entirely on machine A or machine B, or it can be made on both.  The nature of the machines makes their cost functions differ: C_A(x) = 10 + 1/6 x² and C_B(y) = 200 + 1/9 y³.  Total cost is C(x,y) = C_A(x) + C_B(y), where x items are made on machine A and y on machine B.  How many units should be made on each machine in order to minimize total costs if x + y = 10,100 units are required?
Solution:  Set F(x,y,l) = [ 10 + 1/6 x² + 200 + 1/9 y³ ] - l [ x + y - 10100 ]
dF/dx = 1/3 x - l = 0 when l = 1/3 x
dF/dy = 1/3 y² - l = 0 when l = 1/3 y²
so 1/3 x = 1/3 y², i.e., x = y²
dF/dl = -x - y + 10100 = -y² - y + 10100 = 0 when y = -101 or +100.
Since we're making things, -101 doesn't make sense so is discarded
Thus y = 100 and x = 10,000.

---

7.7 - Double Integrals
Example 1: 
Solution: 

ssolved examples

- Antiderivatives
Example 1:  Find an antiderivative of f(x) = 2x.
Solution:  The function F(x) = x² is an antiderivative of f(x) since F ' (x) = 2x = f(x).
Example 2:  Find an antiderivative of f(x) = x⁴.
Solution:  F(x) = 1/5 x⁵ + C
Example 3:  Find an antiderivative of f(x) = 2x + x²
Solution:  F(x) = x² + 1/3 x³ + C
Example 4:  Find an antiderivative of f(x) = x e^x2
Solution:  F(x) = 1/2 e^x2 + C
Example 5:  Evaluate Integral [ x^2/3 ] dx.
Solution:  Integral [ x^2/3 ] dx = [ x^{2/3 + 1} / ( 2/3 + 1 ) ] + C = 3/5 x^5/3 + C
Example 6:  Evaluate Integral [ x^-3 ] dx
Solution:  Integral [ x^-3 ] dx = [ x^-2 / ( -2 ) ] + C = -1/2 x^-2 + C
Example 7:  Evaluate Integral [ 2 x² ] dx
Solution:  Integral [ 2 x² ] dx = 2 Integral [ x² ] dx = 2 [ x³ / 3 + C ] = 2/3 x³ + C
Example 8:  Evaluate Integral [ 2 x^-1/2 + 4x ] dx
Solution:  Integral [ 2x^-1/2 + 4x ] dx = Integral [ 2x^-1/2 ] dx + Integral [ 4x ] dx = 2 Integral [ x^-1/2 ] dx + 4 Integral [ x ] dx = 2 [ x^1/2 / ( 1/2 ) ] + 4 [ x² / 2 ] + C = 4x^1/2 + 2x² + C
Example 9:  Find the antiderivative F(x) of f(x) = 3x for which F(0) = 1
Solution:  F(x) = Integral [ 3x ] dx = 3 Integral [ x ] dx = 3 [ x² / 2 ] + C = 3/2 x² + C.   Now 1 = F(0) = 3/2 ( 0 )² + C = C, so F(x) = 3/2 x² + 1
Example 10:  The marginal cost C_M(x) of producing x kilograms of penicillin is C_M(x) = 2500 + 10x^3/2 dollars per kilogram.   If the start-up cost is $3500, what is the total cost function C(x)?
Solution:  Since C_M(x) = C ' (x), C(x) is an antiderivative of the marginal cost with C(0) = 3500.   Thus C(x) = Integral [ 2500 + 10x^3/2 ] dx = Integral [ 2500 ] dx + Integral [ 10x^3/2 ] dx = 2500 Integral [ 1 ] dx + 10 Integral [ x^3/2 ] dx = 2500 [ x ] + 10 [ 2/5 x^5/2 ] + C.    Now 3500 = C(0) = 2500 ( 0 ) + 4 ( 0 )^5/2 + C = C, so C(x) = 2500 x + 10 x^5/2 + 3500
Example 11:  Evaluate Integral [ 7 / x ] dx
Solution Integral [ 7 / x ] dx = 7 Integral [ 1 / x ] dx = 7 ln | x | + C
Example 12:  Evaluate Integral [ ( x + 1 ) / x ] dx
Solution:  Integral [ ( x + 1 ) / x ] dx = Integral [ x/x + 1/x ] dx = Integral [ 1 ] dx + Integral [ 1 / x ] dx = [ x ] + [ ln | x | ] + C
Example 13:  Evaluate Integral [ 3 e^2x ] dx
Solution:  Integral [ 3 e^2x ] dx = 3 Integral [ e^2x ] dx = 3 [ 1/2 e^2x ] + C = 3/2 e^2x + C
Example 14:  Evaluate Integral [ e^-3x ] dx
Solution:  Integral [ e^-3x ] dx = 1/(-3) e^-3x + C = -1/3 e^-3x + C
Example 15:  Find an antiderivative F(x) of f(x) = 2e^4x with F(0) = 8.
Solution:  F(x) = Integral [ 2 e^4x ] dx = 2 Integral [ e^4x ] dx = 2 [ 1/4 e^4x ] + C = 1/2 e^4x + C.   Then 8 = F(0) = 1/2 e^{4 ( 0 )} + C = 1/2 + C, so C = 15/2 and F(x) = 1/2 e^4x + 15/2

---

6.2, 6.3 - The Definite Integral and Area
Example 1:  Evaluate Integral₀¹ [ x ] dx
Solution:  Integral [ x ] dx = 1/2 x² + C, so Integral₀¹ [ x ] dx = [ 1/2 ( 1 )² + C ] - [ 1/2 ( 0 )² + C ] = 1/2
Example 2:  Evaluate Integral₁³ [ 2 / x ] dx
Solution:  Integral [ 2 / x ] dx = 2 Integral [ 1 / x ] dx = 2 [ ln | x | ] + C, so Integral₁³ [ 2 / x ] dx = [ 2 ln | 3 | + C ] - [ 2 ln | 1 | + C ] = 2 ln 3
Example 3:  Evaluate Integral₀¹ [ x² ] dx
Solution:  Integral [ x² ] dx = 1/3 x³ + C, so Integral₀¹ [ x² ] dx = [ 1/3 ( 1 )³ + C ] - [ 1/3 ( 0 )³ + C ] = 1/3
Example 4:  Evaluate Integral_-1¹ [ x³ ] dx
Solution:  Integral [ x³ ] dx = 1/4 x⁴ + C, so Integral_-1¹ [ x³ ] dx = [ 1/4 ( 1 )⁴ + C ] - [ 1/4 ( -1 )⁴ + C ] = 0
Example 5:  Evaluate Integral₀⁴ [ 2x ( x² + 1 )^1/2 ] dx
Solution:  Integral [ 2x ( x² + 1 )^1/2 ] dx = 2/3 ( x² + 1 )^3/2 + C (check this by taking the derivative!), so Integral₀⁴ [ 2x ( x² + 1 )^1/2 ] dx = [ 2/3 ( ( 4 )² + 1 )^3/2 + C ] - [ 2/3 ( ( 0 )² + 1 )^3/2 + C ] = 2/3 [ ( 17 )^3/2 - 1 ]
Example 6:  Find the area of the region bounded by the parabola y = x², the x-axis, and the vertical lines x = 0 and x = 1
Solution:  Area = Integral₀¹ [ x² ] dx = [ 1/3 ( 1 )³ + C ] - [ 1/3 ( 0 )³ + C ] = 1/3
Example 7:  Find the area of the region bounded by the graph of y = e^2x, the x-axis, and the vertical lines x = -2 and x = 1
Solution:  Area = Integral_-2¹ [ e^2x ] dx = [ 1/2 e^{2 ( 1 )} + C ] - [ 1/2 e^{2 ( -2 )} + C ] = 1/2 [ e² - e^-4 ] = 3.6854
Example 8:  Use a definite integral to find the area of a right triangle of height h and base b
Solution:  Put the right angle of the triangle at the origin, so the triangle is the region to the right of the y-axis, above the x-axis, and below the line connecting the points (0,h) and (b,0).   This last line has slope -h/b, so has equation (using slope-intercept form) y = -h/b x + h.   Then Integral [ -h/b x + h ] dx = -h/b ( 1/2 x² ) + h ( x ) + C.    Therefore Area = Integral₀^b [ -h/b x + h ] dx = [ -h/2b ( b )² + h ( b ) + C ] - [ -h/2b ( 0 )² + h ( 0 ) + C ] = -h/2 ( b ) + h ( b ) = 1/2 hb
Example 9:  Suppose the marginal cost of producing golf balls is given by
C_M(x) = 150 + 0.02 x dollars per gross If 500 gross are already produced, what is the cost of producing the next 500?
Solution:  We're asked to find C(1000) - C(500).   Since total cost is an antiderivative of marginal cost, this difference is equal to Integral₅₀₀¹⁰⁰⁰ [ C_M(x) ] dx = Integral₅₀₀¹⁰⁰⁰ [ 150 + 0.02 x ] dx = [ 150 ( 1000 ) + 0.02 ( 1/2 ( 1000 )² ) + C ] - [ 150 ( 500 ) + 0.02 ( 1/2 ( 500 )² ) + C ] = $82,500
Example 10:  Suppose marginal profit is P_M(x) = 200 - 9 x^1/2.   What is the profit earned from the sale of the 225th item to the 400th item?
Solution:  Profit is an antiderivative of marginal profit, so P(400) - P(225) = Integral₂₂₅⁴⁰⁰ [ 200 - 9 x^1/2 ] dx = [ 200 ( 400 ) - 9 ( 2/3 ( 400 )^3/2 ) + C ] - [ 200 ( 225 ) - 9 ( 2/3 ( 225 )^3/2 ) + C ] = $7,250
Example 11:  Between 1980 and 2000, the rate of oil consumption has risen 1% per year.   In 1980, the rate of consumption was 20 billion barrels per day.   How many barrels of oil were consumed in the 20 years between 1980 and 2000?
Solution:  We first need to find the function expressing the rate of consumption.   We're told that R ' (t) = 0.01 R(t), i.e., that the rate increased by 1% per year.   Thus the rate function is exponential of the form R(t) = R₀ e^{0.01 t}, and R₀ = 20, since the rate in 1980 was 20 billion barrels.   So the total oil consumed is given by Integral₀²⁰ [ R(t) ] dt = Integral₀²⁰ [ 20 e^{0.01 t} ] dt = 20 Integral₀²⁰ [ e^{0.01 t} ] dt = 20 { [ 1/0.01 e^{0.01 ( 20 )} + C ] - [ 1/0.01 e^{0.01 ( 0 )} + C ] } = 20 { 100 e^0.2 - 100 ] = 442.81 billion barrels

---

6.5 - Applications of the Definite Integral
Example 1:  Find the average value of the funtion y = x² between x = 1 and x = 3.
Solution:  [ Integral₁³ x² dx ] / ( 3 - 1 ) = 1/2 [ 1/3 x³ ]₁ ³ = 1/2 [ 1/3 ( 27 ) - 1/3 ( 1 ) ] = 26/6 = 13/3.
Example 2:  The daily output of a assembly line during the month of May is modelled by P(t) = 15 + 40t - t², where t is the t^th work day of the month of May.   What was the average productivity over the course of the 22 work days in May?
Solution:  [ Integral₀²² ( 15 + 40t - t² ) dt ] / ( 22 - 0 ) = 1/22 [ 15t + 20t² - 1/3 t³ ]₀²² = 293.67
Example 3:  The average annual per capita energy consumption (in millions of BTUs) has grown 2% per year since 1940.    In 1940 the average annual per capita energy consumption was 181 million BTUs.   What was the average annual per capita energy consumption between the years 1940 and 1970?
Solution:  Let B(t) = the average annual per capita energy consumption t years after 1940.   Then we want to know the average value of B(t) between t = 0 and t = 30.   First, we need to know what B(t) is.   Since it is growing as a percentage of its present value, i.e., B ' (t) = 0.02 B(t), B(t) is exponential.   Therefore B(t) = B₀ e^{0.02 t} = 181 e^{0.02 t}.   Hence the average value is [ Integral₀³⁰ 181 e^{0.02 t} dt ] / [ 30 - 0 ] = 1/30 [ 181/0.02 e^{0.02 t} ]₀³⁰ = 248 million BTUs.
Example 4:  Find the consumer's surplus for the demand function D(x) = ( x - 5 )² when q = 3.
Solution:  When q = 3, p = D(3) = ( -2 )² = 4, so the consumer's surplus is [ Integral₀³ ( x² - 10x + 25 ) dx ] - [ ( 3 )( 4 ) ] = [ 1/3 x³ - 5 x² + 25 x ]₀³ - 12 = $27.
Example 5:  Find the producer's surplus for the supply function S(x) = x² + x + 3 when q = 3.
Solution:  When q = 3, p = S(3) = 15, so the producer's surplus is [ ( 3 )( 15 ) ] - [ Integral₀³ ( x² + x + 3 ) dx ] = 45 - [ 1/3 x³ + 1/2 x² + 3 x ]₀³ = $22.50.
Example 6:  Suppose the demand function is D(x) = ( x - 5 )² and the supply function is S(x) = x² + x + 3 for a certain item.
a. What is the equilibrium point?
b. What is the consumer's surplus at this point?
c. What is the producer's surplus at this point?
Solution:  a. Set S(x) = D(x).   So x² + x + 3 = x² - 10 x + 25, i.e., 11 x = 22.   Thus q_E = 22/11 = 2 and p_E = S(2) = $9.   Therefore the equilibrium point is (2,$9).
b. Consumer's surplus = [ Integral₀² ( x² - 10x + 25 ) dx ] - [ ( 2 )( 9 ) ] = [ 1/3 x³ - 5 x² + 25 x ]₀² - [ 18 ] = $14.67.
c. Producer's surplus = [ ( 2 )( 9 ) ] - [ Integral₀ ² ( x² + x + 3 ) dx ] = [ 18 ] - [ 1/3 x³ + 1/2 x² + 3 x ]₀² = $7.33.
Example 7:  Suppose money is flowing continuously into a savings account at an annual rate of $1000 per year at an interest rate of 8% compounded continuously.
a. How much money is in the account after 5 years?
b. How much money is in the account after 15 years?
Solution:  a. Integral₀⁵ [ 1000 e^{0.08 t} ] dt = [ 1000/0.08 e^{0.08 t} ]₀ ⁵ = $6147.81.
b. Integral₀¹⁵ [ 1000 e^{0.08 t} ] dt = [ 1000/0.08 e^{0.08 t} ]₀ ¹⁵ = $29,001.46.
Example 8:  Suppose P₀ dollars is invested each year into a savings account paying 8% interest compounded continuously over a period of 20 years.   If we want to have $10,000 in the account at the end of the 20 years, what should P₀ be to ensure this?
Solution:  We want 10000 = Integral₀²⁰ [ P₀ e^{0.08 t} ] dt = P₀ [ 1/0.08 e^{0.08 t} ]₀²⁰ = 49.4 P₀, so we need P₀ = 10000 / 49.4 = $202.38.

---

9.1 - Substitution
Example 1:  If u = x³, find its differential du.
Solution:  du = [ 3 x² ] dx
Example 2:  If u = ln x, find its differential du.
Solution:  du = [ 1 / x ] dx
Example 3:  Evaluate Integral [ 2x e^x2 ] dx
Solution:  Let u = x², so du = 2x dx.  Then Integral [ 2x e^x2 ] dx = Integral e^u du = e^u + C = e^x2 + C.
Example 4:  Evaluate Integral [ 2x / ( 1 + x² ) ] dx
Solution:  Let u = 1 + x², so du = [ 2x ] dx.   Then Integral [ 2x / ( 1 + x² ) ] dx = Integral [ 1 / u ] du = ln | u | + C = ln | 1 + x² | + C
Example 5:  Evaluate Integral [ 3 x² / ( 1 + x³ )² ] dx
Solution:  Let u = 1 + x³, so du = [ 3 x² ] dx.  Then Integral [ 3 x² / ( 1 + x³ )² ] dx = Integral [ 1 / u² ] du = Integral [ u^-2 ] du = u^-1 / ( -1 ) + C = -( 1 + x³ )^-1 + C
Example 6:  Evaluate Integral [ ln( 3x ) / x ] dx.
Solution:  Let u = ln( 3x ), so du = [ 1 / ( 3x ) ]( 3 ) dx = [ 1 / x ] dx.  Then Integral [ ln( 3x ) / x ] dx = Integral [ u ] du = u² / 2 + C = 1/2 ( ln( 3x ) )² + C
Example 7:  Evaluate Integral [ x e^x2 ] dx
Solution:  Let u = x², so du = 2x dx and x dx = 1/2 du.  Then Integral [ x e^x2 ] dx = Integral [ e^u ] ( 1/2 du ) = 1/2 Integral e^u du = 1/2 e^u + C = 1/2 e^x2 + C
Example 8:  Evaluate Integral [ e^x / ( 4 + e^x ) ] dx
Solution:  Let u = 4 + e^x, so du = e^x dx.  Then Integral [ e^x / ( 4 + e^x ) ] dx = Integral [ 1 / u ] du = ln | u | + C = ln | 4 + e^x | + C
Example 9:  Evaluate Integral [ 1 / ( x + 3 ) ] dx
Solution:  Let u = x + 3, so du = dx and Integral [ 1 / ( x + 3 ) ] dx = Integral [ 1 / u ] du = ln | u | + C = ln | x + 3 | + C
Example 10:  Evaluate Integral [ x² ( x³ + 1 )¹⁰ ] dx
Solution:  Let u = x³ + 1, so du = 3x² dx and x² dx = 1/3 du.  Thus Integral [ x² ( x³ + 1 )¹⁰ ] dx = Integral [ u¹⁰ ] ( 1/3 du ) = 1/3 [ u¹¹ / 11 ] + C = 1/33 ( x³ + 1 )¹¹ + C
Example 11:  Evaluate Integral [ x⁴ e^x5 ] dx
Solution:  Let u = x⁵, so du = 5x⁴ dx and x⁴ dx = 1/5 du.  Therefore Integral [ x⁴ e^x5 ] dx = Integral [ e^u ] ( 1/5 du ) = 1/5 [ e^u ] + C = 1/5 e^x5 + C
Example 12:  Evaluate Integral [ 1 / ( x ln( x² ) ) ] dx
Solution:  Let u = ln( x² ), so du = [ 1 / x² ] ( 2x ) dx = [ 2 / x ] dx and 1 / x dx = 1/2 du.  Then Integral [ 1 / ( x ln( x² ) ) ] dx = Integral [ 1 / u ] ( 1/2 du ) = 1/2 [ ln | u | ] + C = 1/2 ln | ln( x² ) | + C
Example 13:  Evaluate Integral [ x ( 4x² + 9 )^1/2 ] dx
Solution:  Let u = 4x² + 9, so du = 8x dx and x dx = 1/8 du.  Then Integral [ x ( 4x² + 9 )^1/2 ] dx = Integral [ u^1/2 ] ( 1/8 du ) = 1/8 [ u^3/2 / ( 3/2 ) ] + C = 1/12 ( 4x² + 9 )^3/2 + C
Example 14:  Evaluate Integral [ ( 7x + 1 )^1/2 ] dx
Solution:  Let u = 7x + 1, so du = 7 dx and dx = 1/7 du.  Then Integral [ ( 7x + 1 )^1/2 ] dx = Integral [ u^1/2 ] ( 1/7 du ) = 1/7 [ u^3/2 / ( 3/2 ) ] + C = 2/21 ( 7x + 1 )^3/2 + C
Example 15:  Evaluate Integral [ x ( x + 1 )^1/2 ] dx
Solution:  Let u = x + 1, so x = u - 1 and dx = du.  Then Integral [ x ( x + 1 )^1/2 ] dx = Integral [ ( u - 1 ) u^1/2 ] du = Integral [ u^3/2 - u^1/2 ] du = u^5/2 / ( 5/2 ) - u^3/2 / ( 3/2 ) + C = 2/5 ( x + 1 )^5/2 - 2/3 ( x + 1 )^3/2 + C

---

9.2 - Integration by Parts
Example 1:  Compute Integral [ x e^x ] dx.
Solution A:  Clearly this is not of the form for any of the standard integrals we have memorized.   Our next idea would be to do a substitution, but letting u = x doesn't do enough and u = e^x doesn't address the problem of the extra x we have in the integrand.    So this problem requires integration by parts, which means we need to decide what to make our f(x) and what should be g(x) in the expression x e^x.   Following hint 1, we select g(x) first and so try g(x) = e^x, in which case f(x) = x.   Then G(x) = e^x and f ' (x) = 1, so that
Integral [ x e^x ] dx = ( x ) ( e^x ) - Integral [ ( 1 ) ( e^x ) ] dx = x e^x - [ e^x + C ] = x e^x - e^x + C Solution B:  Suppose in this example that we had chosen f(x) and g(x) differently as f(x) = e^x and g(x) = x.   Then f ' (x) = e^x while G(x) = x² / 2, so that G(x) is "more complicated" than g(x) was which is our first indication that this might not work.   But if we continue we would have
Integral [ x e^x ] dx = ( e^x ) ( x² / 2 ) - Integral [ ( e^x ) ( x² / 2 ) ] dx, which now leaves us to compute Integral [ 1/2 x² e^x ] dx. However, this isn't a straight-forward integral and is in fact more complicated and "ugly" than what we started with, so with this second clue we would think to try different choices for f(x) and g(x) in our original problem, and thus obtain our solution in part A above. Example 2:  Evaluate Integral [ x ln x ] dx.
Solution:  Let us examine several choices, as follows.
Choice A:  Let f(x) = 1 and g(x) = x ln x.   This will not work because we are back to our original integral, in which we do not know how to integrate g(x) = x ln x.
Choice B:  We let f(x) = x ln x and g(x) = 1.   Then f ' (x) = ( 1 ) ln x + x ( 1 / x ) = ln x + 1 and G(x) = x.   Using the integration by parts formula, we have
Integral [ x ln x ] dx = ( x ln x ) ( x ) - Integral [ ( ln x + 1 ) ( x ) ] dx = x² ln x - Integral [ x ln x + x ] dx = x² ln x - [ Integral [ x ln x ] dx + x² / 2 + C ]
Now we observe that Integral [ x ln x ] dx appears on both sides of the equation, but occurs with a negative sign on the right.   So if we add Integral [ x ln x ] dx to both sides, we have 2 Integral [ x ln x ] dx = x² ln x - 1/2 x² + C
so that (dividing both sides by 2)
Integral [ x ln x ] dx = 1/2 x² ln x - 1/4 x² + C.
This method worked, but a third option below is even easier.   Remember that in math, there is only one right answer, but there might be several ways of getting there! Choice C:  We let f(x) = ln x and g(x) = x.   Then f ' (x) = 1 / x and G(x) = x² / 2, so the integration by parts formula says
Integral [ x ln x ] dx = ( ln x ) ( x² / 2 ) - Integral [ ( 1 / x ) ( x² / 2 ) ] dx = 1/2 x² ln x - Integral [ 1/2 x ] dx = 1/2 x² ln x - 1/2 [ x² / 2 + C ] = 1/2 x² ln x - 1/4 x² + C. Example 3:  Evaluate Integral [ x ( x + 1 )^1/2 ] dx.
Solution:  We let f(x) = x and g(x) = ( x + 1 )^1/2, so f ' (x) = 1 and G(x) = 2/3 ( x + 1 )^3/2.   Then
Integral [ x ( x + 1 )^1/2 ] dx = ( x ) [ 2/3 ( x + 1 )^3/2 ] - Integral [ ( 1 ) ( 2/3 [ x + 1 ]^3/2 ) ] dx = 2/3 x ( x + 1 )^3/2 - 2/3 [ 2/5 ( x + 1 )^5/2 + C ]
= 2/3 x ( x + 1 )^3/2 - 4/15 ( x + 1 )^5/2 + C. Example 4:  Evaluate Integral₁² [ ln x ] dx.
Solution:  First we need to compute Integral [ ln x ] dx.    Let f(x) = ln x and g(x) = 1 so that f ' (x) = 1 / x and G(x) = x.    Then Integral [ ln x ] dx = ( ln x ) ( x ) - Integral [ ( 1 / x ) ( x ) ] dx = x ln x - Integral [ 1 ] dx = x ln x - x + C.   Thus Integral₁² [ ln x ] dx = [ x ln x - x ]₁ ² = [ 2 ln 2 - 2 ] - [ 1 ln 1 - 1 ] = 2 ln 2 - 2 - [ - 1 ] = 2 ln 2 - 1.
Example 5:  Evaluate Integral [ x² e^x ] dx
Solution:  Let f(x) = x² and g(x) = e^x so f ' (x) = 2x and G(x) = e^x.   Then
Integral [ x² e^x ] dx = ( x² ) ( e^x ) - Integral [ ( 2x ) ( e^x ) ] dx = x² e^x - 2 Integral [ x e^x ] dx.
Now we can either recall from Example 1 what Integral [ x e^x ] dx is or we may use integration by parts a second time, with f(x) = x and g(x) = e^x (so f ' (x) = 1 and G(x) = e^x) to obtain Integral [ x² e^x ] dx = x² e^x - 2 [ x e^x - e^x + C ] = x² e^x - 2x e^x + 2 e^x + C.

---

9.6, 12.2 - Improper Integrals and Probability
Example 1:  Find the area of the region under the graph of y = 1 / x² over the interval [1,infinity).
Solution:  Integral₁^infinity [ x^-2 ] dx = lim_{b -> infinity} Integral₁ ^b [ x^-2 ] dx = lim_{b -> infinity} [ -x^-1 ]₁^b = lim_{b -> infinity} [ ( -1 / b ) + ( 1 ) ] = 1
Example 2:  Find the area of the region under the graph of y = 1 / x over the interval [1,infinity).
Solution:  Integral₁^infinity [ x^-1 ] dx = lim_{b -> infinity} Integral₁ ^b [ x^-1 ] dx = lim_{b -> infinity} [ ln | x | ]₁^b = lim_{b -> infinity} [ ( ln b ) - ( ln 1 ) ] does not exist, since ln x -> infinity as x -> infinity.
Example 3:  Evaluate Integral₀ ^infinity [ 2 e^{-2 x} ] dx
Solution:  Integral₀^infinity [ 2 e^{-2 x} ] dx = lim_{b -> infinity} Integral₀ ^b [ 2 e^{-2 x} ] dx = lim_{b -> infinity} [ -e^{-2 x} ]₀^b = lim_{b -> infinity} [ ( -e^-2b ) - ( -e⁰ ) ] = 0 + 1 = 1.
Example 4:  Evaluate Integral_-infinity ^infinity [ x e^-x2 ] dx
Solution:  Integral_-infinity^infinity [ x e^-x2 ] dx = Integral_-infinity ⁰ [ x e^-x2 ] dx + Integral₀ ^infinity [ x e^-x2 ] dx = lim_{a -> -infinity} Integral_a⁰ [ x e^-x2 ] dx + lim_{b -> infinity} Integral₀^b [ x e^-x2 ] dx = lim_{a -> -infinity} [ -1/2 e^-x2 ]_a⁰ + lim_{b -> infinity} [ -1/2 e^-x2 ]₀^b = lim_{a -> -infinity} [ ( -1/2 ) - ( -1/2 e^-a2 ) ] + lim_{b -> infinity} [ ( -1/2 e^-b2 ) - ( -1/2 ) ] = [ -1/2 + 0 ] + [ 0 + 1/2 ] = 0.
Example 5:  What is the probability of drawing an ace from a well-shuffled deck of playing cards?
Solution:  There are a total of 4 aces out of 52 cards, so the probability is 4/52 = 1/13 = 0.077.
Example 6:  Suppose that x is the arrival time of a bus at a bus stop between 2pm and 5pm.  What is the probability that the bus arrives between 4pm and 5pm?
Solution:  Suppose the bus is as likely to arrive at one moment as it is to arrive at another, so x is uniformly distributed and f(x) = 1 / ( 5 - 2 ) = 1/3.   Then P( 4 < x < 5 ) = Integral₄⁵ [ 1/3 ] dx = [ 1/3 x ]₄⁵ = 5/3 - 4/3 = 1/3.
Example 7:  Verify that f(x) = 3/117 x² is a probability density function for x on the interval [2,5].
Solution:  Clearly f(x) > 0 on the interval, so we only need to show that 1 = Integral₂⁵ [ 3/117 x² ] dx = 3/117 [ 1/3 x³ ]₂⁵ = 1/117 [ 125 - 8 ] = 1, as desired.
Example 8:  A company produces transistors.  It determines that the life-span t of a transistor is between 3 and 6 years with a probability density function f(t) = 24 t^-3.
a.  Find P( t < 4 )
b.  Find P( 4 < t < 5 )
Solution:  a. P( t < 4 ) = P( 3 < t < 4 ) = Integral₃⁴ [ 24 t^-3 ] dt = 24 [ t^-2 / ( -2 ) ]₃⁴ = 0.58
b. P( 4 < t < 5 ) = Integral₄⁵ [ 24 t^-3 ] dt = 24 [ t^-2 / ( -2 ) ]₄ ⁵ = 0.27
Example 9:  Find k so that f(x) = k x² is a probability density function for x on [2,5].
Solution:  We need 1 = Integral₂⁵ [ k x² ] dx = k [ x³ / 3 ]₂⁵ = k [ 125/3 - 8/3 ] = 117/3 k.   Thus k = 3/117.
Example 10:  The distance x between successive cars on a highway has probability density function f(x) = k e^{-k x}, for 0 < x < infinity, where k = 1/a and a = average distance between successive cars.   If a = 166 feet, what is the probability that x is less than 50 feet?
Solution:  P( x < 50 ) = Integral₀⁵⁰ [ 1/166 e^{-1/166 x} ] dx = [ -e^{-1/166 x} ]₀⁵⁰ = 0.26
Example 11:  Given the probability density function f(x) = 1/2 x for x on the interval [0,2], find the mean (expected value) of x.
Solution:  m = Integral₀ ² [ x ^. ( 1/2 x ) ] dx = 1/2 Integral₀ ² [ x² ] dx = 1/2 [ x³ / 3 ]₀² = 1/2 [ 8/3 - 0 ] = 8/6 = 4/3

---

7.1, 7.2 - Functions of Several Variables and Partial Derivatives
Example 0:  Functions of several variables are used often in our everyday lives. For instance, in the winter we are very interested in the wind chill factor, which uses both the actual, raw temperature outside and the current wind speed to tell us how cold it actually feels. In the summer, the heat index combines the temperature and relative humidity to measure how hot it really feels outside.
Example 1:  A company makes two items, guns and butter.   It sells each gun for $4 and each pound of butter for $6.  Find the company's total revenue function and evaluate it when it sells 25 guns and 10 pounds of butter.
Solution:  R(x,y) = 4x + 6y, where x = # guns and y = pounds of butter.  So R(25,10) = 4(25) + 6(10) = 160.
Example 2:  The total cost of a company, in thousands of dollars, is given by C(x,y,z,w) = 4x² + 5y + z - ln( w + 1 ), where x = money spent on labor, y = money spent on raw materials, z = money spent on advertising, and w = money spent on machinery.  Find C(3,2,0,10).
Solution:  C(3,2,0,10) = 4(3)² + 5(2) + (0) - ln( (10) + 1 ) = 36 + 10 - ln 11 = 43.6 thousands of dollars.
Example 3:  The wind chill at temperature T and wind speed v is given by W(T,v) = 91.4 - [ ( 10.45 + 6.68 v^1/2 - 0.477 v )( 457 - 5 T ) ] / 110, where T is the current temperature in degrees Fahrenheit and v is the wind speed in miles per hour.
a.  Find the wind chill when T = 30^o F, v = 20 mph.
b.  Find the wind chill when T = 20^o F, v = 20 mph.
c.  Find the wind chill when T = 20^o F, v = 40 mph.
Solution:  a. W(30,20) = 5.49
b. W(20,20) = -8.51
c. W(20,40) = -17.71
Example 4:  Find the first order partial derivatives of f(x,y) = x²y³ + xy + 4y².
Solution:  df/dx = (2x)y³ + (1)y + 0 = 2xy³ + y and
df/dy = x²(3y²) + x(1) + 8y = 3x²y² + x + 8y.
Example 5:  Find the first order partial derivatives of f(x,y,z) = x² - xy + y² + 2yz + 2z² + z.
Solution:  df/dx = 2x - y
df/dy = -x + 2y + 2z
df/dz = 2y + 4z + 1
Example 6:  Find the first order partial derivatives of f(x,y) = 3x²y + xy.
Solution:  df/dx = 3(2x)y + (1)y = 6xy + y
df/dy = 3x²(1) + x(1) = 3x² + x
Example 7:  Find the first order partial derivatives of f(x,y) = e^xy + y ln x
Solution:  df/dx = e^xy[(1)y] + y( 1/x ) = y e^xy + y/x
df/dy = e^xy[x(1)] + (1)ln x = x e^xy + ln x
Example 8:  Find the marginal productivity of labor for the Cobb-Douglas production function p(x,y) = 50 x^2/3 y^1/3, where x = # units of labor and y = # units of capital, when x = 125 and y = 64.
Solution:  The marginal productivity of labor is dp/dx = 50(2/3 x^-1/3)y^1/3, so dp/dx (125,64) = 100/3 (125)^-1/3 (64)^1/3 = 80/3
Example 9:  Find d² f / dy dx, where f(x,y) = 3xy² + 2xy + x².
Solution:  df/dx = 3(1)y² + 2(1)y + 2x = 3y² + 2y + 2x, so d² f / dy dx = d/dy [ df/dx ] = d/dy [ 3y² + 2y + 2x ] = 6y + 2
Example 10:  Find all second order partial derivatives of f(x,y) = x² y³ + x⁴ y + x e^y.
Solution:  df/dx = (2x)y³ + (4x³)y + (1)e^y and df/dy = x²(3y²) + x⁴(1) + x e^y.  Thus
d²f/dx² = d/dx [ 2xy³ + 4x³y + e^y ] = 2(1)y³ + 4(3x²)y + 0 = 2y³ + 12x²y
d²f/dydx = d/dy [ 2xy³ + 4x³y + e^y ] = 2x(3y²) + 4x³(1) + e^{y = 6xy2 + 4x3 + ey}
d²f/dxdy = d/dx [ 3x²y² + x⁴ + x e^y ] = 3(2x)y² + 4x³ + (1)e^y = 6xy² + 4x³ + e^y = d²f/dydx
d²f/dy² = d/dy [ 3x²y² + x⁴ + x e^y ] = 3x²(2y) + 0 + x(e^y) = 6x²y + x e^y

---

7.3 - Maxima and Minima of Functions of Two Variables
Example 1:  Find all first and second order partial derivatives of f(x,y) = 5x²y - 7xy + 2y.
Solution:  df/dx = 5(2x)y - 7(1)y + 0 = 10xy - 7y
df/dy = 5x²(1) - 7x(1) + 2(1) = 5x² - 7x + 2
d²f/dx² = 10(1)y - 0 = 10y
d²f/dydx = 10x(1) - 7(1) = 10x - 7
d²f/dxdy = 5(2x) - 7(1) + 0 = 10x - 7
d²f/dy² = 0 - 0 + 0 = 0
Example 2:  Use the First Derivative Test to locate the relative minimum of f(x,y) = 3x² - 4xy + 3y² + 8x - 17y + 30.
Solution:  df/dx = 6x - 4y + 8 = 0 when 4y = 6x + 8 or y = 3/2 x + 2.
df/dy = -4x + 6y - 17 = -4x + 6(3/2 x + 2) - 17 = -4x + 9x + 12 - 17 = 5x - 5 = 0 when x = 1.
When x = 1, we also have y = 3/2 + 2 = 7/2.  So f has its relative minimum at the point (1,7/2).
Example 3:  Use the First Derivative Test to locate the relative minimum of f(x,y) = 1/2 x² + y² - 3x + 2y - 5.
Solution:  df/dx = x - 3 = 0 when x = 3
df/dy = 2y + 2 = 0 when y = -1
So f has its relative minimum at the point (3,-1).
Example 4:  Use the First and Second Derivative Tests to find the relative maxima and/or minima of f(x,y) = y³ - x² + 6x - 12y + 5.
Solution:  df/dx = -2x + 6 = 0 when x = 3
df/dy = 3y² - 12 = 3(y² - 4) = 3(y - 2)(y + 2) = 0 when y = +2
So the possible extreme points are at (3,2) and (3,-2).
Now d²f/dx² = -2
d²f/dydx = 0
d²f/dxdy = 0
d²f/dy² = 6y
So D(x,y) = (d²f/dx²)(d²f/dy²) - (d²f/dydx)²
At (3,2), we have D(3,2) = (-2)[6(2)] = -24 < 0, so NEITHER
At (3,-2), we have D(3,-2) = (-2)[6(-2)] = +24 > 0 and d²f/dx² = -2 < 0, so RELATIVE MAXIMUM
Example 5:  Use the First Derivative Test to locate the relative minimum of f(x,y) = -x² - 8xy - y².
Solution:  df/dx = -2x - 8y = 0 when x = -4y
df/dy = -8x - 2y = -8(-4y) - 2y = 30y = 0 when y = 0.
So the only possible extreme point is when x = -4(0) = 0 and y = 0.
Now d²f/dx² = -2
d²f/dydx = -8
d²f/dxdy = -8
d²f/dy² = -2
So D(x,y) = (-2)(-2) - (-8)² = 4 - 64 = -60 < 0
so f has NEITHER a relative max or min at (0,0).
Example 6:  A firm makes two kinds of golf balls.  One sells for $3 each and the other sells for $2 each.  Thus the revenue from selling x (thousand) balls at $3 and y (thousand) balls at $2 is
R(x,y) = 3x + 2y The company has determined that the total cost function of producing x thousand balls of the first type and y thousand balls of the second is given by
C(x,y) = 2x² - 2xy + y² - 9x + 6y + 7 Find the number of balls of each type that should be made in order to maximize this company's profits. Solution:  P(x,y) = R(x,y) - C(x,y) = [ 3x + 2y ] - [ 2x² - 2xy + y² - 9x + 6y + 7 ] = -2x² + 2xy - y² + 12x - 4y - 7.  Now
dP/dx = -4x + 2y + 12 = 0 when 2y = 4x - 12
dP/dy = 2x - 2y - 4 = 0 when 2y = 2x - 4
so both are zero when 4x - 12 = 2x - 4, i.e., when 2x = 8, so x = 4 and y = 1/2 [ 2(4) - 4 ] = 2
Thus our possible max occurs at the point (4,2).
Now d²P/dx² = -4, d²P/dydx = 2, and d²P/dy² = -2, so
D(x,y) = ( -4 )( -2 ) - ( 2 )² = 8 - 4 = 4.
Thus D(4,2) = 4 > 0 and d²P/dx² (4,2) = -4 < 0
so profit is maximized when x = 4 (thousand) and y = 2 (thousand).

---

7.4 - Lagrange Multipliers and Constrained Optimization
Lecture 1
Example 0:  The autumn demand for home insulation depends on the unit cost x for heating fuel and the unit cost y for insulation.  Suppose the demand for insulation is given by
d(x,y) = -4x² + 2xy - 6y² + 28x + 62y + 850 For what values of x and y is the demand maximum?
Solution:  dd/dx = -8x + 2y + 28 = 0 when y = 4x - 14
dd/dy = 2x - 12y + 62 = 2x - 12( 4x - 14 ) + 62 = -46x + 230 = 0
when x = 5 and y = 4( 5 ) - 14 = 6
Now d²d/dx² = -8, d²d/dydx = 2, and d²d/dy² = -12
So D(x,y) = ( -8 )( -12 ) - ( 2 )² = 92
Since D(5,6) = 92 > 0 and d²d/dx² (5,6) = -8 < 0
demand is maximum when x = 5 and y = 6.
Example 1:  Maximize f(x,y) = 2xy subject to the constraint g(x,y) = 2x + 3y - 6 = 0.
Solution:  Set F(x,y,l) = f(x,y) - l g(x,y) = [ 2xy ] - l [ 2x + 3y - 6 ]
dF/dx = 2y - 2l = 0 when l = y
dF/dy = 2x - 3l = 0 when l = 2/3 x
So y = l = 2/3 x, which we plug into
dF/dl = -2x - 3y + 6 = -2x -3( 2/3 x ) + 6 = -4x + 6 = 0
when x = 3/2, so y = 2/3 ( 3/2 ) = 1.
So f is maximize at ( 3/2 , 1 ), with value f(3/2,1) = 3
Example 2:  Minimize f(x,y) = x² + y² subject to g(x,y) = xy - 2 = 0.
Solution:  Set F(x,y,l) = [ x² + y² ] - l [ xy - 2 ]
dF/dx = 2x - yl = 0 when l = 2x/y
dF/dy = 2y - xl = 0 when l = 2y/x
so 2x/y = 2y/x, i.e., x² = y² (by cross multiplying)
so y = +x, which we now plug into
dF/dl = -xy + 2 = -x² + 2 = 0 when y = x
or x² + 2 = 0 when y = -x, but this can't happen so y = x.
So y = x = +2^1/2, making the possible minimimum points at either ( 2^1/2,2^1/2 ) or ( -2^1/2,-2^1/2 )
Now f(+2^1/2,+2^1/2) = 4 in both cases, so both are minima.
Example 3:  Maximize P(r,s) = r - s subject to r² + s² = 4.
Solution:  Set F(r,s,l) = [ r - s ] - l [ r² + s² - 4 ]
dF/dr = 1 - 2rl = 0 when l = 1/2r
dF/ds = -1 - 2sl = 0 when l = -1/2s
so 2s = -2r, i.e., s = -r, which we now use in
dF/dl = -r² - s² + 4 = -r² - ( -r )² + 4 = -2r² + 4 = 0
when r = +2^1/2 and s = +2^1/2
Now we want to maximize, so consider
P( 2^1/2,-2^1/2 ) = 2^3/2
P( -2^1/2,2^1/2 ) = -2^3/2
so P is maximized at ( 2^1/2,-2^1/2 )
Example 4:  Minimize and maximize f(x,y) = 2x² + 3y² subject to x⁴ + y² = 1.
Solution:  Set F(x,y,l) = [ 2x² + 3y² ] - l [ x⁴ + y² - 1 ]
dF/dx = 4x - 4x³ l = 4x ( 1 - x² l ) = 0 when x = 0 or l = 1/x²
dF/dy = 6y - 2y l = 2y ( 3 - l ) = 0 when y = 0 or l = 3
When x = 0, 0⁴ + y² = 1 implies y = +1
when y = 0, x⁴ + 0² = 1 implies x = +1
otherwise 3 = l = 1/x² so x = +( 1/3 )^1/2 and y = +( 8/9 )^1/2
Now f(0,+1) = 3
f(+1,0) = 2
f(+( 1/3 )^1/2,+( 8/9 )^1/2) = 2/3 + 8/3 = 10/3
so the max is at (+( 1/3 )^1/2,+( 8/9 )^1/2) and the min is at (+1,0).
Example 5:  Minimize x² + 2xy + 2y² + x + y subject to 1 - x - 2y = 0.
Solution:  Set F(x,y,l) = [ x² + 2xy + 2y² + x + y ] - l [ 1 - x - 2y ]
dF/dx = 2x + 2y + 1 + l = 0 when l = -1 - 2y - 2x
dF/dy = 2x + 4y + 1 + 2l = 0 when l = -1/2 - x - 2y
so -1 - 2y - 2x = -1/2 - x - 2y, so -x = 1/2 or x = -1/2
dF/dl = -1 + x + 2y = -1 + ( -1/2 ) + 2y = -3/2 + 2y = 0 when y = 3/4
so the minimum occurs when x = -1/2 and y = 3/4
Lecture 2
Example 1:  Maximize f(x,y) = 4 - x² - y² subject to x + 2y = 10.
Solution:  Set F(x,y,l) = [ 4 - x² - y² ] - l [ x + 2y - 10 ]
dF/dx = -2x - l = 0 when l = -2x
dF/dy = -2y - 2l = 0 when l = -y, so y = 2x
dF/dl = -x - 2y + 10 = -5x + 10 = 0 when x = 2 and y = 4
Example 2:  Suppose the production function of a company is given by f(x,y) = 8 x^1/4 y^3/4, where x is the number of units of labor and y is the number of units of capital.  If x + y = 64, what is the maximum productivity?
Solution:  Set F(x,y,l) = [ 8 x^1/4 y^3/4 ] - l [ x + y - 64 ]
dF/dx = 2 x^-3/4 y^3/4 - l = 0 when l = 2 x^-3/4 y^3/4
dF/dy = 6 x^1/4 y^-1/4 - l = 0 when l = 6 x^1/4 y^-1/4
so 2 x^-3/4 y^3/4 = 6 x^1/4 y^-1/4, i.e., y = 3x
dF/dl = - x - y + 64 = - 4x + 64 = 0 when x = 16 and y = 48
then the productivity is f(16,48) = 291.78
Example 3:  The production function of (another) company is f(x,y) = 4 x^1/2 y^1/2, where x is the number of units of labor and y the number of units of capital.  If labor costs $300 per unit and capital costs $100 per unit and the firm can only invest $18,000, what values of x and y maximize productivity?
Solution:  Set F(x,y,l) = [ 4 x^1/2 y^1/2 ] - l [ 300x + 100y - 18000 ]
dF/dx = 2 x^-1/2 y^1/2 - 300l = 0 when l = 1/150 x^-1/2 y^1/2
dF/dy = 2 x^1/2 y^-1/2 - 100l = 0 when l = 1/50 x^1/2 y^-1/2
so 1/150 x^-1/2 y^1/2 = 1/50 x^1/2 y^-1/2, i.e., y = 3x
dF/dl = -300x - 100y + 18000 = -600x + 18000 = 0 when x = 30 and y = 90.
Example 4:  The total sales S of a one-product firm are given by S(L,M) = 2ML - L², where M is the cost of materials and L is the cost of labor.  Find the maximum sales subject to the budget constraint M + L = 60.
Solution:  Set F(L,M,l) = [ 2ML - L² ] - l [ M + L - 60 ]
dF/dL = 2M - 2L - l = 0 when l = 2M - 2L
dF/dM = 2L - l = 0 when l = 2L
so 2L = 2M - 2L, i.e., M = 2L
dF/dl = -M - L + 60 = -3L + 60 = 0 when L = 20 and M = 40.
Example 5:  A product can be made entirely on machine A or machine B, or it can be made on both.  The nature of the machines makes their cost functions differ: C_A(x) = 10 + 1/6 x² and C_B(y) = 200 + 1/9 y³.  Total cost is C(x,y) = C_A(x) + C_B(y), where x items are made on machine A and y on machine B.  How many units should be made on each machine in order to minimize total costs if x + y = 10,100 units are required?
Solution:  Set F(x,y,l) = [ 10 + 1/6 x² + 200 + 1/9 y³ ] - l [ x + y - 10100 ]
dF/dx = 1/3 x - l = 0 when l = 1/3 x
dF/dy = 1/3 y² - l = 0 when l = 1/3 y²
so 1/3 x = 1/3 y², i.e., x = y²
dF/dl = -x - y + 10100 = -y² - y + 10100 = 0 when y = -101 or +100.
Since we're making things, -101 doesn't make sense so is discarded
Thus y = 100 and x = 10,000.

---

7.7 - Double Integrals
Example 1: 
Solution: 

differential equation introduce

Solving Differential Equationswith Integrating Factors
mccp-dobson-0111
Introduction
Suppose we have the first order differential equation
dy
dx
+ Py = Q
where P and Q are functions involving x only. For example
dy
dx
+
3y
x
=
ex
x3
or
dy
dx
−
3y
x + 1
= (x + 1)4.
We can solve these differential equations using the technique of an integrating factor.
Integrating Factor
We multiply both sides of the differential equation by the integrating factor I which is defined as
I = eR P dx.
General Solution
Multiplying our original differential equation by I we get that
dy
dx
+ Py = Q , I
dy
dx
+ IPy = IQ
, Z (I
dy
dx
+ IPy) dx = Z IQdx
, Iy = Z IQdx since
d
dx
(Iy) = I
dy
dx
+ IPy by the product rule.
As both I and Q are functions involving only x in most of the problems you are likely to meet, R IQdx
can usually be found. So the general solution to the differential equation is found by integrating IQ
and then re-arranging the formula to make y the subject.
Example
To find the general solution of the differential equation
dy
dx
+
3y
x
=
ex
x3
www.mathcentre.ac.uk

c Katy Dobson
University of Leeds
Alan Slomson
University of Leeds
mathcentre community project
encouraging academics to share maths support resources
All mccp resources are released under an Attribution Non-commerical Share Alike licence
we first find the integrating factor
I = eR P dx = eR 3
x
dx
now Z 3
x
dx = 3 ln x = ln x3
hence I = eln x3
= x3.
Then we multiply the differential equation by I to get
x3 dy
dx
+ 3x2y = ex
so integrating both sides we have x3y = ex + c where c is a constant. Thus the general solution is
y =
ex + c
x3
.
Example
To find the general solution of the differential equation
dy
dx
−
3y
x + 1
= (x + 1)4
we first find the integrating factor
I = eR P dx = eR −3
x+1
dx
now Z −3
x + 1
dx = −3 ln(x + 1) = ln(x + 1)−3
hence I = eln(x+1)−3
= (x + 1)−3 =
1
(x + 1)3
.
Then multiplying the differential equation by I we get
1
(x + 1)3
dy
dx
−
3y
(x + 1)4 = (x + 1)
so integrating both sides we have
y
(x + 1)3 =
1
2
x2 + x + c where c is a constant.
Thus the general solution is
y = (x + 1)3(
1
2
x2 + x + c).
Exercises
Find the general solution of
1.
dy
dx
+
2y
x
=
sin x
x2 2.
dy
dx
−
y
x
= −xe−x 3.
dy
dx
+ 2xy = x 4.
dy
dx
−
2y
x
= 3x3
Answers
1. y =
c − cos x
x2 2. y = x(e−x + c) 3. y =
1
2
+ ce−x2
4. y =
3
2
x4 + cx2
www.mathcentre.ac.uk

c Katy Dobson
University of Leeds
Alan Slomson
University of Leeds

integration solutions

Math 331.5: Homework 2
Solutions
1. Find the general solution of each di erential equation and draw several representative solutions.
(i) dy
dt = 􀀀y + 5
First divide by the right hand side.
1
􀀀y + 5
dy
dt
= 1
Next integrate with respect to t.
(1)
Z
1
􀀀y + 5
dy
dt

dt =
Z
1 dt
Now use u-substitution for the integral on the left hand side. Let u = 􀀀y + 5 so du =
􀀀dy
dt dt. The integral becomes
Z
1
􀀀y + 5
dy
dt

dt = 􀀀
Z
1
u
du
= 􀀀ln juj + C
= 􀀀ln j5 􀀀 yj + C
Then (1) becomes
􀀀ln j5 􀀀 yj = t + C
ln j5 􀀀 yj = 􀀀t + C
To solve for y, we exponentiate both sides.
eln j5􀀀yj = e􀀀t+C
j5 􀀀 yj = e􀀀teC
5 􀀀 y = eCe􀀀t
5 􀀀 y = Ce􀀀t
􀀀y = Ce􀀀t 􀀀 5
y = Ce􀀀t + 5
(ii) dy
dt = 2y + 5
Following the method outlined above, the general solution is y = Ce2t 􀀀 5
2 .
2
-20 -15 -10 -5 0 5 10 15 20
-12
-8
-4
4
8
12
Figure 1. y = Ce􀀀t + 5
-10 -7.5 -5 -2.5 0 2.5 5 7.5 10
-5
-2.5
2.5
5
Figure 2. y = Ce2t 􀀀 5
2
3
2. Solve the following initial value problems and draw the solutions.
(i) dy
dt = 3y 􀀀 7, y(0) = 1.
First nd the general solution as in problem 1. Divide by the right hand side.
1
3y 􀀀 7
dy
dt
= 1
Next integrate with respect to t.
(2)
Z
1
3y 􀀀 7
dy
dt

dt =
Z
1 dt
Now use u-substitution for the integral on the left hand side. Let u = 3y􀀀7 so du = 3dy
dt dt.
The integral becomes
Z
1
3y 􀀀 7
dy
dt

dt =
1
3
Z
1
u
du
=
1
3
ln juj + C
=
1
3
ln j3y 􀀀 7j + C
Then (2) becomes
1
3
ln j3y 􀀀 7j = t + C
ln j3y 􀀀 7j = 3t + C
To solve for y, we exponentiate both sides.
eln j3y􀀀7j = e3t+C
j3y 􀀀 7j = e3teC
3y 􀀀 7 = eCe3t
3y 􀀀 7 = Ce3t
3y = Ce3t + 7
y = Ce3t +
7
3
Now use the initial condition, y(0) = 1, to nd C. Setting t = 0 and y = 1,
1 = Ce3(0) +
7
3
C = 􀀀
4
3
So, the desired solution is
y = 􀀀
4
3e3t +
7
3
(ii) dy
dt = y 􀀀 1, y(0) = 1.
Observe that y = 1 () dy
dt = 0, so y = 1 is an equilibrium solution. Then the desired
solution to the ODE is y(t) = 1.
4
-10 -7.5 -5 -2.5 0 2.5 5 7.5 10
-5
-2.5
2.5
5
Figure 3. y = 􀀀4
3 e3t + 7
3
-5 -4 -3 -2 -1 0 1 2 3 4 5
-3
-2
-1
1
2
3
Figure 4. y = 1
5
3. Suppose an object with mass 10 kg and air drag coe cient 2 kg/s is dropped from above the
ground.
(i) Find the time that must elapse for the object to reach 95% of its terminal velocity.
The di erential equation for velocity is dv
dt = 9:8 􀀀 1
5v. Since the object is dropped, the
initial condition is v(0) = 0. In class, we saw that the solution for this initial value problem
is
v(t) = 􀀀49e􀀀0:2t + 49:
Letting t ! 1 in the above, the terminal velocity is 49 m/s. So, 95% of the terminal
velocity is 46.55 m/s. We wish to nd the time T when v(T) = 46:55. Substituting into
the equation,
46:55 = 􀀀49e􀀀0:2T + 49
􀀀2:45 = 􀀀49e􀀀0:2T
0:05 = e􀀀0:2T
ln(0:05) = ln(e􀀀0:2T )
ln(0:05) = 􀀀0:2T
T = 􀀀5 ln(0:05) 14:98
Therefore, the object reaches 95% of its terminal velocity after about 15 seconds.
(ii) What is the minimal height so that the object hits the ground at this speed?
We need to nd the equation for the displacement, x. Since dx
dt = v, the di erential
equation for displacement is dx
dt = 􀀀49e􀀀0:2t + 49. The rate function depends only on the
independent variable, t, so we can integrate to nd x.
Z
dx
dt
dt =
Z
(􀀀49e􀀀0:2t + 49)dt
x(t) = 245e􀀀0:2t + 49t + C
To nd C, we use the initial condition x(0) = 0. Substituting t = 0 and x = 0, we get
C = 􀀀245. So, our equation is
x(t) = 245e􀀀0:2t + 49t 􀀀 245
Setting t = 􀀀5 ln(0:05),
x = 245eln(0:05) + 49(􀀀5 ln(0:05)) 􀀀 245 501
Therefore, the minimal height is about 501 meters above ground.
4. Calculate the impact velocity if you jump o a 7 foot wall (ignore air drag).
Ignoring air drag (setting
 = 0), the di erential equation for velocity is dv
dt = 32:2, where g =
32:2ft/s2 is the acceleration due to gravity. Integrating,
Z
dv
dt
dt =
Z
32:2 dt
v(t) = 32:2t + C
6
Now use the initial condition v(0) = 0 to nd C. Substituting t = 0 and v = 0 we nd C = 0.
Then the equation for velocity is v(t) = 32:2t.
To compute the impact velocity, we need to know the time of impact. So, we nd the equation for
displacement, x. The di erential equation for displacement is dx
dt = v(t) = 32:2t. Integrating,
Z
dx
dt
dt =
Z
32:2t dt
x(t) = 16:1t2 + C
The initial displacement is x(0) = 0, so substituting t = 0 and x = 0 in the equation, we get C = 0.
Then the equation for displacement is x(t) = 16:1t2. Now we need to nd T when we hit the
ground, i.e. x = 7.
7 = 16:1T2
T =
p
7=16:1
We want T > 0, so we take the positive square root, T =
p
7=16:1. Putting this value back into
the equation for velocity, the impact velocity is v = 32:2
p
7=16:1 21:2ft/s.
5. The population of the United States was 8.6 million in 1820 and 40 million in 1897. Suppose
that the population increases at a rate proportional to the current population, with growth rate r.
(i) Write a di erential equation for the population of the U.S.
Let P = P(t) be the population of the U.S. in millions t years after 1820. Let r be the
growth rate (constant of proportionality). Then the di erential equation for P is dP
dt = rP.
(ii) Find the solution which satis es the conditions above.
First nd the general solution as in problem 1. Divide by the right hand side.
1
rP
dP
dt
= 1
Next integrate with respect to t.
(3)
Z
1
rP
dP
dt

dt =
Z
1 dt
Now use u-substitution for the integral on the left hand side. Let u = rP so du = r dP
dt dt.
The integral becomes
Z
1
rP
dP
dt

dt =
1
r
Z
1
u
du
=
1
r
ln juj + C
=
1
r
ln jrPj + C
Then (3) becomes
1
r
ln jrPj = t + C
ln jrPj = rt + C
7
To solve for P, we exponentiate both sides.
eln jrPj = ert+C
jrPj = erteC
rP = eCert
rP = Cert
P = Cert
Now use the initial condition, P(0) = 8:6, to nd C. Setting t = 0 and P = 8:6,
8:6 = Cer(0)
C = 8:6
Then P(t) = 8:6ert. To nd r we use the information P(77) = 40.
40 = 8:6e77r
40=8:6 = e77r
ln(40=8:6) = ln(e77r)
ln(40=8:6) = 77r
r =
ln(40=8:6)
77 0:02
So, P(t) = 8:6e0:02t.
(iii) Use this model to calculate the population of the U.S. in 2003. If the actual population of
the U.S. in 2003 was 291 million, how good is this estimate?
P(183) = 8:6e0:02(183) 332
(iv) According to this model, when would the population of the U.S. double?
We wish to nd the time T when the population is twice its initial value, i.e. P(T) =
2P(0) = 17:2.
17:2 = 8:6e0:02T
2 = e0:02T
ln 2 = ln(e0:02T )
0:02T = ln 2
T = (ln 2)=0:02 35
Therefore, according to this model, the population of the US doubled by 1855.
8
6. Suppose that a building loses heat in accordance with Newton's law of cooling (see homework
1, problem 8) and that the rate constant is 0:15=hr. Assume that the interior temperature is 70 F
when the heating system fails. If the external temperature is 10 F, how long will it take for the
interior temperature to fall to 32 F?
Let T be the temperature of the building in F and t be time in hours. The surrounding temperature
is 10 F, so by Newton's law, the di erential equation for T is
dT
dt
= 0:15(10 􀀀 T):
The building starts o at 70 F before the heating system fails, so rst we need to solve the initial
value problem
(4)
(
dT
dt
= 0:15(10 􀀀 T)
T(0) = 70
Solving this IVP by the method outlined in problem 2, we nd T = 10 + 60e􀀀0:15t.
To nd how long it will take for the temperature to fall to 32 F, we set T = 32 and solve for
t.
32 = 10 + 60e􀀀0:15t
22 = 60e􀀀0:15t
22=60 = e􀀀0:15t
ln(22=60) = ln
􀀀
e􀀀0:15t
ln(22=60) = 􀀀0:15t
t =
ln(22=60)
􀀀0:15 6:69
So, the building reaches freezing in about 6.7 hours.
7. Consider an electric circuit containing a capacitor, resistor and battery. The charge Q(t) on the
capacitor satis es the equation
(5) R
dQ
dt
+ Q
C
= V
where R is the resistance, C is the capacitance, and V is the constant voltage supplied by the
battery.
(i) If Q(0) = 0, nd an expression for Q in terms of the constants R;C and V .
We need to solve the initial value problem
(6)
(
R
dQ
dt
+ Q
C
= V
Q(0) = 0
First nd the general solution. We will rewrite the equation as
dQ
dt
= V
R
􀀀
Q
CR
9
Next we divide by the right hand side.
1
V
R 􀀀 Q
CR
dQ
dt
= 1
Next integrate with respect to t.
(7)
Z 
1
V
R 􀀀 Q
CR
dQ
dt
!
dt =
Z
1 dt
Now use u-substitution for the integral on the left hand side. Let u = V
R 􀀀 Q
CR so du =
􀀀 1
CR
dQ
dt dt. The integral becomes
Z 
1
V
R 􀀀 Q
CR
dQ
dt
!
dt = 􀀀CR
Z
1
u
du
= 􀀀CRln juj + K
= 􀀀CRln

V
R
􀀀
Q
CR

+ K
Then (7) becomes
􀀀CRln

V
R
􀀀
Q
CR

= t + K
ln

V
R
􀀀
Q
CR

= 􀀀
1
CR
t + K
To solve for Q, we exponentiate both sides.
elnj V
R􀀀 Q
CRj = e􀀀 1
CRt+K

V
R
􀀀
Q
CR

= e􀀀 1
CRteK
V
R
􀀀
Q
CR
= eKe􀀀 1
CRt
V
R
􀀀
Q
CR
= Ke􀀀 1
CRt
􀀀
Q
CR
= Ke􀀀 1
CRt 􀀀
V
R
Q = Ke􀀀 1
CRt + CV
Now use the initial condition, Q(0) = 0, to nd K. Setting t = 0 and Q = 0,
0 = Ke0 + CV
K = 􀀀CV
Then Q(t) = 􀀀CV e􀀀 1
CRt + CV .
(ii) Find the limiting value QL that Q(t) approaches as t ! 1.
As t ! 1, e􀀀t=CR ! 0. So,
QL = lim
t!1
Q(t) = CV:
10
(iii) Suppose that at some time T the battery is removed and the circuit is closed again. If
Q(T) = QL, nd Q(t) and sketch its graph.
V is the voltage supplied by the battery, so if the battery is removed, the voltage is zero.
Then for t > T, the di erential equation for Q becomes
(8) R
dQ
dt
+ Q
C
= 0
At time T, Q(T) = QL = CV . So, we want to solve the IVP
(9)
(
R
dQ
dt
+ Q
C
= 0
Q(T) = CV
From part i, the general solution of (8) is
Q = Ke􀀀 1
CRt:
Using the initial condition Q(T) = CV , we nd
CV = Ke􀀀 1
CRT
or
K = CV e
1
CRT
Then for t > T,
Q(t) = CV e
1
CRT e􀀀 1
CRt = CV e(T􀀀t)=CR:
Now note that Q(t) = CV is an equilibrium solution of the original di erential equation
(5). So, if Q(T) = CV at time T, we must have had Q(t) = CV for all times t up to T.
Hence, the solution Q with this behavior is de ned by
Q(t) =

CV if t T
CV e(T􀀀t)=CR if t T
0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 4.4 4.8
0.25