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 ex2
Solution: F(x) = 1/2 ex2 + C
Example 5: Evaluate Integral [ x2/3 ] dx.
Solution: Integral [ x2/3 ] dx = [ x2/3 + 1 / ( 2/3 + 1 ) ] + C = 3/5 x5/3 + 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 [ x1/2 / ( 1/2 ) ] + 4 [ x2 / 2 ] + C = 4x1/2 + 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 CM(x) of producing x kilograms of penicillin is CM(x) = 2500 + 10x3/2 dollars per kilogram. If the start-up cost is $3500, what is the total cost function C(x)?
Solution: Since CM(x) = C ' (x), C(x) is an antiderivative of the marginal cost with C(0) = 3500. Thus C(x) = Integral [ 2500 + 10x3/2 ] dx = Integral [ 2500 ] dx + Integral [ 10x3/2 ] dx = 2500 Integral [ 1 ] dx + 10 Integral [ x3/2 ] dx = 2500 [ x ] + 10 [ 2/5 x5/2 ] + C. Now 3500 = C(0) = 2500 ( 0 ) + 4 ( 0 )5/2 + C = C, so C(x) = 2500 x + 10 x5/2 + 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 e2x ] dx
Solution: Integral [ 3 e2x ] dx = 3 Integral [ e2x ] dx = 3 [ 1/2 e2x ] + C = 3/2 e2x + 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) = 2e4x with F(0) = 8.
Solution: F(x) = Integral [ 2 e4x ] dx = 2 Integral [ e4x ] dx = 2 [ 1/4 e4x ] + C = 1/2 e4x + C. Then 8 = F(0) = 1/2 e4 ( 0 ) + C = 1/2 + C, so C = 15/2 and F(x) = 1/2 e4x + 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 = e2x, the x-axis, and the vertical lines x = -2 and x = 1
Solution: Area = Integral-21 [ e2x ] dx = [ 1/2 e2 ( 1 ) + C ] - [ 1/2 e2 ( -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 = Integral0b [ -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
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 [ CM(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 PM(x) = 200 - 9 x1/2. 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 x1/2 ] 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 e0.01 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 e0.01 t ] dt = 20 Integral020 [ e0.01 t ] dt = 20 { [ 1/0.01 e0.01 ( 20 ) + C ] - [ 1/0.01 e0.01 ( 0 ) + C ] } = 20 { 100 e0.2 - 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 tth 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 e0.02 t = 181 e0.02 t. Hence the average value is [ Integral030 181 e0.02 t dt ] / [ 30 - 0 ] = 1/30 [ 181/0.02 e0.02 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 qE = 22/11 = 2 and pE = 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 e0.08 t ] dt = [ 1000/0.08 e0.08 t ]0 5 = $6147.81.
b. Integral015 [ 1000 e0.08 t ] dt = [ 1000/0.08 e0.08 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 e0.08 t ] dt = P0 [ 1/0.08 e0.08 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 ex2 ] dx
Solution: Let u = x2, so du = 2x dx. Then Integral [ 2x ex2 ] dx = Integral eu du = eu + C = ex2 + 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 ex2 ] dx
Solution: Let u = x2, so du = 2x dx and x dx = 1/2 du. Then Integral [ x ex2 ] dx = Integral [ eu ] ( 1/2 du ) = 1/2 Integral eu du = 1/2 eu + C = 1/2 ex2 + C
Example 8: Evaluate Integral [ ex / ( 4 + ex ) ] dx
Solution: Let u = 4 + ex, so du = ex dx. Then Integral [ ex / ( 4 + ex ) ] dx = Integral [ 1 / u ] du = ln | u | + C = ln | 4 + ex | + 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 ex5 ] dx
Solution: Let u = x5, so du = 5x4 dx and x4 dx = 1/5 du. Therefore Integral [ x4 ex5 ] dx = Integral [ eu ] ( 1/5 du ) = 1/5 [ eu ] + C = 1/5 ex5 + 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 [ u1/2 ] ( 1/8 du ) = 1/8 [ u3/2 / ( 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 [ u1/2 ] ( 1/7 du ) = 1/7 [ u3/2 / ( 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 ) u1/2 ] du = Integral [ u3/2 - u1/2 ] du = u5/2 / ( 5/2 ) - u3/2 / ( 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 ex ] 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 = ex 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 ex. Following hint 1, we select g(x) first and so try g(x) = ex, in which case f(x) = x. Then G(x) = ex and f ' (x) = 1, so that
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
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
so that (dividing both sides by 2)
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
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
= 2/3 x ( x + 1 )3/2 - 4/15 ( x + 1 )5/2 + C.
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 ex ] dx
Solution: Let f(x) = x2 and g(x) = ex so f ' (x) = 2x and G(x) = ex. Then
Now we can either recall from Example 1 what Integral [ x ex ] dx is or we may use integration by parts a second time, with f(x) = x and g(x) = ex (so f ' (x) = 1 and G(x) = ex) to obtain
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: Integral1infinity [ x-2 ] dx = limb -> infinity Integral1 b [ x-2 ] dx = limb -> infinity [ -x-1 ]1b = limb -> 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: Integral1infinity [ x-1 ] dx = limb -> infinity Integral1 b [ x-1 ] dx = limb -> infinity [ ln | x | ]1b = limb -> 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: Integral0infinity [ 2 e-2 x ] dx = limb -> infinity Integral0 b [ 2 e-2 x ] dx = limb -> infinity [ -e-2 x ]0b = limb -> infinity [ ( -e-2b ) - ( -e0 ) ] = 0 + 1 = 1.
Example 4: Evaluate Integral-infinity infinity [ x e-x2 ] dx
Solution: Integral-infinityinfinity [ x e-x2 ] dx = Integral-infinity 0 [ x e-x2 ] dx + Integral0 infinity [ x e-x2 ] dx = lima -> -infinity Integrala0 [ x e-x2 ] dx + limb -> infinity Integral0b [ x e-x2 ] dx = lima -> -infinity [ -1/2 e-x2 ]a0 + limb -> infinity [ -1/2 e-x2 ]0b = lima -> -infinity [ ( -1/2 ) - ( -1/2 e-a2 ) ] + limb -> 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 v1/2 - 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 = 30o F, v = 20 mph.
b. Find the wind chill when T = 20o F, v = 20 mph.
c. Find the wind chill when T = 20o 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) = exy + y ln x
Solution: df/dx = exy[(1)y] + y( 1/x ) = y exy + y/x
df/dy = exy[x(1)] + (1)ln x = x exy + ln x
Example 8: Find the marginal productivity of labor for the Cobb-Douglas production function p(x,y) = 50 x2/3 y1/3, 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)y1/3, 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 ey.
Solution: df/dx = (2x)y3 + (4x3)y + (1)ey and df/dy = x2(3y2) + x4(1) + x ey. Thus
d2f/dx2 = d/dx [ 2xy3 + 4x3y + ey ] = 2(1)y3 + 4(3x2)y + 0 = 2y3 + 12x2y
d2f/dydx = d/dy [ 2xy3 + 4x3y + ey ] = 2x(3y2) + 4x3(1) + ey = 6xy2 + 4x3 + ey
d2f/dxdy = d/dx [ 3x2y2 + x4 + x ey ] = 3(2x)y2 + 4x3 + (1)ey = 6xy2 + 4x3 + ey = d2f/dydx
d2f/dy2 = d/dy [ 3x2y2 + x4 + x ey ] = 3x2(2y) + 0 + x(ey) = 6x2y + x ey
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
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
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 = +21/2, making the possible minimimum points at either ( 21/2,21/2 ) or ( -21/2,-21/2 )
Now f(+21/2,+21/2) = 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 = +21/2 and s = +21/2
Now we want to maximize, so consider
P( 21/2,-21/2 ) = 23/2
P( -21/2,21/2 ) = -23/2
so P is maximized at ( 21/2,-21/2 )
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 x1/4 y3/4, 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 x1/4 y3/4 ] - l [ x + y - 64 ]
dF/dx = 2 x-3/4 y3/4 - l = 0 when l = 2 x-3/4 y3/4
dF/dy = 6 x1/4 y-1/4 - l = 0 when l = 6 x1/4 y-1/4
so 2 x-3/4 y3/4 = 6 x1/4 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 x1/2 y1/2, 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 x1/2 y1/2 ] - l [ 300x + 100y - 18000 ]
dF/dx = 2 x-1/2 y1/2 - 300l = 0 when l = 1/150 x-1/2 y1/2
dF/dy = 2 x1/2 y-1/2 - 100l = 0 when l = 1/50 x1/2 y-1/2
so 1/150 x-1/2 y1/2 = 1/50 x1/2 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: CA(x) = 10 + 1/6 x2 and CB(y) = 200 + 1/9 y3. Total cost is C(x,y) = CA(x) + CB(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:
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