Calculus for Business, Economics, Life Sciences, and Social Sciences, 13E Barnett, Ziegler, Byleen S

Calculus

Chapter 5

In Problems 5–8, find each indefinite integral.

5. 3

2

+2x-8)dx

òx3(x2+2x-8)dx

æ3 6. òç èx2

æ3 òç èx2

= ò( x5 + 2 x4 - 8 x3 ) dx =

x6 2 x5 8 x4 +C + 5 4 6

=

x6 2x5 2x4 +C + 6 5

5ö x ÷ dx ø

x )dx (3x-2 -

5ö x ÷ dx = ò ø

5

2

3x-1 x7 2 = - 7 +C -1 2 =-

3 x

-

2 x7 +C 7

æ5x + 3 ö 7. òç ÷ dx è x3 ø æ5 3ö æ5x + 3 ö dx = òçèx2 + x3 ÷dx òç ÷ è x3 ø ø

( 5 x2 3)dx

= ò -x 23 +-

5 x- 1 3 x+ +C -1 -2 3 5 =- +C x 2x2

=

For exam and test prep, contact excellentessaywriters@gmail.com 5-3 ..

Calculus

Chapter 5

æ5x5 +3x 6e xö 8. òç ÷dx x6 è ø

æ5x5 +3x6exö æ5 ö dx =òç + 3e x ÷dx òç ÷ x6 è x è ø ø =5lnx+ 3ex + C

In Problems 9–12, find the particular antiderivative of each derivative that satisfies the given conditions.

9. R'(x)=8x3+9x2+2; R(1)=16

First find the indefinite integral of the function.

R(x) = ò(8x3 + 9x 2 +2) dx 8 x4 9x3 2 x1 + + C R(x)= + 3 4 1 +2 R(x)= 2 x4 + 3 3 x +C x C. Using the condition given, find the specific value of 2 +C R( x) = 2x 4 + x 3 3x + 16 = 2(1)4 + 3(1)3 + 2(1)C+ 16 = 2 + 3 + 2 + C 16 = 7 + C 9=C Substituting the specific value ofC yields the particular equation R(x)=2x4+3x3+2x+9.

For exam and test prep, contact excellentessaywriters@gmail.com 5-4 ..

Calculus Chapter 5

dy 10. =5et+4t-8; y(0)=8 dt First find the indefinite integral of the function.

dy = (5et + 4t - 8)dt òdy = ò(5et + 4t - 8) dt y = 5et +

4t2 8t 1 2 - 1 +C

y = 5et + 2t2 - 8t + C C. Using the condition given, find the specific value of

y =5e t+ 2t 2- 8t + C 0 8 =5e + 2(0)2 - 8(0) + C 8 = 5 +0 + 0 + C 8=5+C 3=C

Substituting the specific value ofC yields the particular equation y=5et+2t2-8t+3.

5-5 ..

Calculus

11.

Chapter 5

dD 5 x2 - 9 ; = dx x2

D(9)=50

First find the indefinite integral of the function.

æ5x2 - 9 ö dD = ç ÷dx x2 ø è 9 2 )dx dD = ( 5 - xòdD = ò(5- 9x- 2) dx 5 1 x- 1 D(x) = x - 9 +C -1 1 9 D(x) = 5x + + C x C. Using the condition given, find the specific value of

9 D( x) = 5 x + + C x 9 50 = 5(9) + + C 9 50 = 45 +1 + C 50 = 46 + C 4 =C

9 Substituting the specific value ofC yields the particular equationD(x) = 5 x + + 4. x

5-6 ..

Calculus

Chapter 5

12. h'(x)=6x-1+7x-2; h(1)=3 First find the indefinite integral of the function.

h'(x) = 6 x-1 + 7 x - 2 h(x) =ò( 6 +7x-2) dx x

h(x) = 6ln x -

7x-1

+C -1 7 h(x) = 6ln x + + C x C. Using the condition given, find the specific value of

7 h( x) = 6ln x + x + C 7 3 = 6ln1 + + C 1 3 = 6(0) + 7 + C 3 =7 +C -4 = C Substituting the specific value ofC yields the particular equation 7 h(x)=6lnx+ x -4.

5-7 ..

Calculus

Chapter 5

5-8 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-2 Integration by Substitution Goal: To find the indefinite integrals using general indefinite integral formulas

Formulas: General Indefinite Integral Formulas 1 .

2 .

4 3 . .

5

) = ò[ f (x)]n f '( xdx

[f(x)]n+ 1

+C, n+1 f(x) f '(x)d x = e f (x) + C òe 1 ò f '( x)dx = ln f ( x) + C f(x) +1 un unòdu = n¹ -1 +C, n +1 euòdu = eu + C 1 ò du = ln u +C u

n¹ -1

.

Definition: Differentials 6 If y=f(x)defines a differentiable function, then .1. The differential dx of the independent variable x is an arbitrary real number.

2. The differential dy of the dependent variable y is defined as the product of f'(x)and dx: dy=f'(x)dx Procedure :

Integration by Substitution

1. Select a substitution that appears to simplify the integrand. In particular, try to select u so the du is a factor in the integrand.

2. Express the integrand entirely in terms of u and du, completely eliminating the original variable and its differential.

3. Evaluate the new integral if possible.

4. Express the antiderivative found in step 3 in terms of the original variable.

5-9 ..

Calculus

Chapter 5

In Problems 1–8, find each indefinite integral and check the result by differentiating.

3 1. ò(32x + 7 x-2)(6x+7)dx

Let u=3x2+7x-2,therefore du=(6x+7)dx.Rewrite the original integral in terms of the variable u and solve.

y = ò(3x2 + 7x-2)3(6x+7)dx y = u3òdu y=

u4

+C 4 1 y = (3x 2 + x7 - 2)4 + C 4 Check:

1 y = 4 (3x2 +7x- 2)4 +C 1 dy d = 4 (4)(3x2 +7x- 2)3 (3x2 + 7 x - 2) dx dx dy (3x2 +7x- 2)3(6x+ = 7) dx 2. 5ò (3x4 +8x-2)(12x3+8)dx

Let u= 3x4+8x-2,therefore du = (12x3 +8)dx.Rewrite the original integral in terms of the variable uand solve. y = ò5 (3x4 +8x - 2)(12 x3 +8) dx 1 y = ò(3x4 +8 x - 2) 5(12 x3 +8) dx 1

y = òu5du 6

y=

u 5 6 5

+C

6 5 y = 6 (3x4 +8 x - 2 ) 5+ C

55

y = 6 (3x4 +8x - 2)6 + C 5-10 ..

Calculus

Chapter 5

Check: 5 y = 6 5 (3x4 +8x- 2)6 + C 5 6 y = 6 (3x4 +8x - 2) 5+ C 1 dy 5æ6ö x4 d = ç ÷(3 +8x - 2) 5 (3x4 +8 x - 2) dx 6 è 5ø dx dy (3x4 +8x - 2) 13(12x3 = +8) dx dy 5 = (3x4 +8x- 2)(12x3 +8) dx

t3+2 - 1 3. ò 2 t 4 + 1 6 t dt

Let u=2t4+16t-1,therefore du = (8t3 +16)dt Þ du = 8(t3 + 2)dt.Rewrite the original integral in terms of the variable and u solve.

1 8(t3 +2) dt y= ò 4 8 2t +16t- 1 1 1 y = ò du 8 u 1 y = ln u +C 8 ln 2t4 +16t - 1 y= +C 8 Check:

y=

ln2t 4 +16t- 1

+C 8 1 d (2t4 +16t dy -1) = 4 dt 8(2t +16t - 1) dx dy 1 (8t 3 + 16) = dt 8(2t4 +16t - 1) dy t3 + 2 = dt 2t4 +16t - 1 5-11 ..

Calculus

Chapter 5

4. -òe0.9xdx

Let u=-0.9x,therefore du=-0.9dx.Rewrite the original integral in terms of the variable u and solve.

y = òe- 0.9 x dx 1 e-0.9x y=(- 0.9dx) ò 0.9 1 euò du y=0.1 1 eu+C y=0.9 10 y = - e - 0.9x +C 9 Check:

e- 0.9 x + C y = 10 9 d

-

10 0.9 x d e(-0.9x) 9 dx

y 10 d = = - 9 e - 0.9x (-0.9) x d - -0.9x =e y d x 5. òx(x+7)7dx d y Let d u=x+7,therefore du=dx.If we rewrite the original integral in terms of the variable u, the substitution would not be complete. We also need u-7=x.Now rewrite the x original integral in terms of u and solve.

y = òx(x + 7)7 dx y = ò(u - 7)u7 du y = ò(u8- 7u7) du u9 7u8 y=9 +C 8 (x + 7)9 7( x +7)8 +C y= 8 9 5-12 ..

Calculus

Chapter 5

Check: (x+7)9 7(x+7)8 y= +C 9 8 dy 1 (9)(x+7)8 d 7 d = (x + 7 ) - 8 (8)( x + 7)7 dx (x + 7) dx dt 9 dy (x+7)8 - 7(x+7)7 = dt dy (x+7)7(x+7- 7) = dt dy = x(x+7)7 dt

2(ln(3x2 ))4 dx 6. ò x Let u=ln(3x2 ),therefore du = 6 x dx Þ du = 2x dx.Rewrite the original integral in 3 x2 terms of the variable uand solve.

2(ln(3x2 ))4 dx y=ò x y = òu4du

y

u5 = 5+

C 5

x ) (ln3 y= +C 2

5

Check: 5

(ln 3x ) +C y= 2

5

dy 1 4d (ln 3x2 ) = (5)(ln 3x2 ) dt 5 dx x dy (ln 3 2 )4 6 = x dt 3 x2 dy 2(ln 3x2 )4 dt = x 5-13 ..

Calculus

7. ò

Chapter 5

1 - 1 x5 dx e x6

Let u=-x-5,therefore du=5x-6dxÞdu=5dx.Rewrite the original integral in x6

terms of the variable u and solve.

1 - 1 x5 dx y = ò6 e x 1 - 1 x5 5 × dx y = òe 5 x6 y = 1 òeu du 5 1 u y = e +C 5 1 - 1 x5 y= e +C 5 Check:

1 - 1 y = e x5 + C 5 dy 1 - 1 x5 d (-x- 5 ) = e dt 5 dx dy 1 - 1 x5 (5x-6) = e dt 5 - 1 dy e x5 dt = x6

5-14 ..

Calculus

8.

Chapter 5

dy =12x2(2x3+7)5Þ

dy=12x2(2x3+7)5dx

dx

2 Let u =x3+7,therefore variable u and solve.

du=6x2 dx.Rewrite the original integral in terms of the

y = ò12x2(2x3+ 7)5 dx y = ò2(2x3 + 7)5 6x2dx y = 2 u5 ò du 2u6 y = 6 +C 1 y = (2x3 + 7)6 + C 3 Check:

1 y = 3 (2x3 +7)6 +C 1 dy d = 3 (6)(2x3 +7)5 (2x3 + 7) dx dt 2(2x3 +7)5(6 dy = x2 ) dt dy =12x2(2x3 +7)5 dt

9. The indefinite integral can be found in more than one way. Given the integral, ò2x(x2+3)2dx,first use the substitution method to find the indefinite integral and then find it without using substitution.

Using substitution, let u=x2+3,therefore du = 2xdx. Rewrite the original integral in terms of the variable u and solve.

y = 2ò x (x 2 + 3)2dx y = u2ò du u3 +C 3 1 + y = 3 ( x2 +3)3 C y=

5-15 ..

Calculus

Chapter 5

Expanding the final function yields the following:

1 y = 3 ( x2 +3)3 + C 1 y = 3 ( x2 + 3)(x2 + 3)(x2 + 3) + C 1 y = 3 ( x6 + 9x 4 + 2 x7 2+ 27) + C x 6 y = + 3x 4 +x9 2C+ 3 Note: The “new” value of C above is the addition of the constant 9 and the original constant C from the integration.

Without using substitution to find the given indefinite integral, you first would multiply the integrand as follows

2x(x2+3)2 = 2 x( x2 + 3)(x2 +3)

= 2 x( x4 + 6x2 +9) = 2 x5 +12x3 +18x Now find the indefinite integral using the above function as the integrand.

+ + 12x3 18x) dx y = ò(2x5 2x6 12x4 18x2 + + = +C y 6 4 2 x6 + + 9x2 C y = + 3x4 3

5-16 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-3 Differential Equations; Growth and Decay Goal: To solve differential equations that involve growth and decay.

Theorem 1 :

Exponential Growth Law

dQ =rQand Q(0 )=Q 0,then Q=Q0ert , dt where Q0 = amount of Q at t=0 r = relative growth rate (expressed as a decimal) t = time Q = quantity at time t If

Table 1:

Exponential Growth

Description Unlimited growth

Model

Solution y=cekt

dy dt =ky

k,t>0 y(0) = c Exponential decay

dy dt =-ky

y = ce-kt

k,t>0 y(0)=c Limited growth

dy dt =k(M-y)

y=M(1-e-

dy = ky(M - y) dt

y=

kt

k, t > 0 y(0)=0 Logistic Growth

k, t > 0 M y(0) = 1+c

5-17 ..

M

1+ ce- kMt

)

Calculus

Chapter 5

In Problems 1–4, find the general or particular solution, as indicated, for each differential equation.

1.

dy =8x-5 dx

The differential equation can be found by using the integration properties from Section 6.1.

y = ò8x- 5 dx y=

8x- 4

+C -4 y = - 2x- 4 + C

d y(0)=9 =- 104e2x5 x - ; y d The x differential equation can be found by using the integration properties from Section 6.2. Let u=-2x5,therefore du = - 10 x 4 dx.

Rewrite the original integral in terms of the variable u and solve.

2.

y =ò- 10 x4-2x5dx e y = òeudu y = eu + C y = e- 2x5+ C C. Using the condition given, find the specific value of x5

y = e-2 + C 5

9 = e-2(0) + C 9 = e0 + C 9 =1 + C 8 =C 5

Substituting the specific value ofC yields the particular equationy = e- 2 x +8.

5-18 ..

Calculus

3.

dy

Chapter 5

=-8y

dx The differential equation is in the form of exponential decay; therefore, using the exponential decay model in Table 1 yield’s the following solution:

y=ce-8x

4.

dx

=3x;

x(0)=6

dt The differential equation is in the form of unlimited growth; therefore, using the unlimited growth model in Table 1 yield’s the following solution:

x=ce3t Using the condition given, find the specific value of C.

x = ce3t 3(0)

6 = ce 0 6 = ce 6 =c

Substituting the specific value of C yields the particular solution x=6e3t.

5. Find the amount A in an account after t years if

dA =0.02A dt

and

A(0)=8000

The exponential growth law (unlimited growth) applies to the situation. Since we are given the initial amount, the amount in an account after t years would be:

A=8000e0.02t

5-19 ..

Calculus

Chapter 5

6. A single injection o a drug is administered to a patient. The amount Q in the body decreases at a rate proportional to the amount present. For a particular drug, the rate dQ is 6% per hour. Thus, =-0.06Qand Q(0)=Q 0where t is time in hours. dt a. If the initial injection is 5 milliliters [Q(0)=5], find Q=Q(t)satisfying both conditions. b. How many milliliters (to two decimal places) are in the body after 8 hours? c. How many hours (to two decimal places) will it take for half the drug to be left in the body? a. By the exponential growth law (exponential decay) for the given conditions, we have the equation Q(t)=5e-0.06t.

b. After 8 hours, t=8,therefore

Q(8 = 5e- 0.06(8) Q) = 5e- 0.48 Q(8 = 5(0.61878) Q » 3.09 ) (8 Therefore, after 8 hours, approximately 3.09 milliliters will remain in the body. ) (8 the initial injection is 2.5 milliliters. We need to find t such that c.

Half Q(t)=2.5. ) Q( t ) = 5e 0.06 2 . 5 =- t 0.06 e- 0.06t 0 . 5 = 5e . 5e-t 0.06t l n 0= -ln

ln 0.5= 0.06t 11.55»- t Therefore, it will take approximately 11.55 hours for the amount of the drug to be half of the initial amount.

5-20 ..

Calculus

Chapter 5

7. A company is trying to expose a new product to as many people as possible through radio ads. Suppose that the rate of exposure to new people is proportional to the number of those who have not heard of the product out of L possible listeners. No one is aware of the product at the start of the campaign, and after 15 days, 50% of L dN are aware of the product. Mathematically, =k(L-N), N(0)=0,and dt N(15)=0.5L. a. Solve the differential equation. b. What percent of L will have been exposed after 7 days of the campaign? c. How many days will it take to expose 75% of L? a. By the exponential growth law (limited growth) for the given conditions, we have the equation

N(t) = L(10.5L = e-

kt

)

15k )

0.5 =1L(1- e- 15k -0.5 =-e-e- 15k 0.5 = e- 15k ln 0.5= lne- 15k ln 0.5 = 15k - k 0.046 = Therefore, the solution isN( t) = L(1- e - 0.046t ).

b. After 7 days, t=7,therefore

N(7) = L(1- e-0.046(7)) N(7) = L(1- 0.724698) N(7) = L(0.275302) N(7) » 0.275L Therefore, after 7 days, approximately 27.5% of L will have been exposed to the product.

5-21 ..

Calculus c.

Chapter 5

Find t such that N(t)=0.75L.

N(t) = L( - e -0.046 ) t ) 0.75 L= 1 - e -0.046

0.75 =1L( - e- 0.046t t - 0.25 = 1- e- 0.046t 0.25 = e- 0.046t ln 0.25= lne- 0.046t ln 0.25= - 0.046t 30.1 » t Therefore, it will take approximately 30 days to expose 75% ofto L the new product.

5-22 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-4 The Definite Integral Goal: To calculate the values of definite integrals using the properties.

Theorem :

Limits of Left and Right Sums

If f(x)>0and is either increasing or decreasing on [a, b], then its left and right sums n®¥. approach the same real number as Theorem:

Limit of Riemann Sums

If f is a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I asn ® ¥ .

Definition:

Definite Integral

Let f be a continuous function on [a, b]. The limit I of Riemann sums for f on [a, b] is b called the definite integral of f from a to b and is denotedaf(x)dx. as ò

Properties:

Definite Integrals

a 1. òaf(x)dx=0 2.

3. b a 4.

5. òaf(x)dx=-òbf(x)dx

b b òakf(x)dx=kòaf(x)dx, k a constant b b b òa[f(x)±g(x)]dx=òaf(x)dx±òag(x)dx b c b òaf(x)dx=òaf(x)dx+òcf(x)dx

5-23 ..

Calculus

Chapter 5

Sn for the function f(x)=13-x3.

In Problems 1 and 2, calculate the indicated Riemann sum 1. Partition [–1, 9] into five subintervals of equal length, and for each subinterval [ xk-1, xk ],let ck =( xk-1+ xk )/2.

x =

c1 =

9- (-1) 10 =5 =2 5

- 1+1 =0 2

c2 =

f (0) =13- (0)3 f (0) =13

c3 =

f (2) =13- (2)3 f (2) = 5

5+7 =6 2

c4 =

1 +3 =2 2

c5 =

3 +5 =4 2

f (4) =13- (4)3 f (4) =- 51

7+9 =8 2

f (6) =13- (6)3 f (6) =- 203

f (8 =13 - (8)3 f ) =- 499 (8 S5 = f (c1) ×Dx + f (c2 )) ×Dx f c+ 3)(×D+ x f (c 4)×D+ x f (c5)×Dx S5 = (13)(2) + (5)(2) + (- 51)(2) + ( -203)(2) + ( -499)(2)

S5 = 26+10- 102- 406- 998 S5 =- 1470 2. Partition [–4, 8] into four subintervals of equal length, and for each subinterval [ xk-1, xk ],let ck =(2 xk-1+ xk )/3.

x =

c1 =

8- (-4) 12 = =3 4 4

2(- 4) + (- 1) = -3 3

f (-3) =17 - 2(- 3)2 f (-3) =- 1

c2 =

2(- 1) + 2 =0 3

f (0) =17- 2( 0) 2 f (0) =17

5-24 ..

Calculus

Chapter 5

2(2)+5 =3 c3= 3 f(3)=17-2(3)2 f(3)=-1

c4=

2(5)+8 =6 3

f(6)=17-2(6)2 f(6)=-55

S4 = fc1 ( ) ×x + fc2 ) ×x+ fc3 ( ) ×x+f(c4)×x S4 = (-1)(3) + (17)(3)+(-1)(3)+(-55)(3) ( S4 = 3 + 51 - 3 -165 S4 - 120 = In Problems 3–7, calculate the definite integral, given that 42

4 ò0xdx=8

ò0x dx=

279 7 2 ò4 x dx = 3

64 3

4 3. ò03xdx

4 4 ò03xdx = 3ò0 xdx = 3(8) 4 ò03xdx = 24

4 4. ò0 (2x + x2 )dx 4 4 2 + ò04 x2 dx ò0(2x + x )dx = 2 0xdx ò 64 = 2(8) + 3 64 =16 + 3 112 4 2 ò0(2x + x )dx = 3

5-25 ..

Calculus

Chapter 5

5. 7ò2x2 dx 0 7 2 7 2 ò02 x dx = 2 ò0x dx

=

2ò0x( 4dx + ò x dx) 7 2 4

2

æ64 279 ö + çè 3 ÷ 3 ø æ343ö =2 ÷ çè 3 ø 686 7 2 ò02 x dx = 3 =2

0 7xdx 6. ò4

4 0 ò47xdx = 7ò0 xdx - 7(8) 0 ò57xdx = 56 -

= 7. ò77( x2 +2x-15)dx Since the upper and lower limits of integration are the same, the value of the integral would be ò77 (2 x +2x-15)dx=0.

5-26 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-5 The Fundamental Theorem of Calculus Goal: To use the fundamental theorem of calculus to solve problems.

Theorem 1 :

Fundamental Theorem of Calculus

If f is a continuous function on [a, b], and F is any antiderivative of f, then b

ò af(x)dx=F(b)-F(a) Definition:

Average Value of a Continuous Function f over [a, b]

1 b òaf(x)dx b-a

In Problems 1–7, evaluate the integrals.

5 1. ò1(8x3+3)dx 5

æ8x4 3 x ö + ÷ ò1(8x + 3) dx = ç 1ø è 4 5

3

1

=

5

(2x4 +3x) 1

=( 2(5)4 + 3(5)) - (2(1)4 + 3 (1) ) =1265 5 5 3 ò1(8x + 3) dx =1260

5-27 ..

Calculus

Chapter 5

2. 3ò(2ex )dx 0 3

3 x x ò0(2e )dx = 2e 2e0

= 2e 3 - 2e 0 3 = 38.17 -2 3 x ò0(2e )dx »

2 3 3. ò2(2x +5x 2 -x+27)dx

Since the upper limit and lower limit of integration are the same value, the definite 2 integral value will be 2(2x3+5x2-x+27)dx=0. ò

5 4. 22ò dx 13 x+3 22

ò13

5 -1 22 dx = ò13 5(x + 3) 2 dx x+3 1

=

22

5(x + 3) 2 1

2

13 1 22

=10(x +3) 2

13 1

22

ò13

5 x+3

1

=10(22 + 3) 2 - 10(13+3) 2 =10(5) - 10(4) = dx 10

5-28 ..

Calculus

Chapter 5

5. ò0123x(3x 4 +2) 6 dx We will solve the problem using the substitution method of integration. Let u=3x4+2,therefore du=12x3dxÞdu=6(2x3dx).Rewrite the original integral in terms of the variable u and solve.

1 x=1 6 1 3 4 6 ò02 x (3x + 2) dx = 6 òx=0 u du u7 x=1 =

=

42

x=0

(3x4 +2)7

1

42 0 (3(1)4 + 2)7 (3(0)4 +2)7 = 42 42 1 3 4 2)6 dx » 1857.07 ò02 x (3x +

3 x2 - 2 x+ 5 dx 6. ò2 1 x3 - x2 +5x- 3

We will solve the problem using the substitution method of integration. Let u=x3-x2+5x-3,therefore du=(3x2-2x+5)dx.Rewrite the original integral in terms of the variable u and solve.

3 x2 - 2x 5+ 1 dx = òxx==12 du ò1 x3- x2 + 5x - 3 u 2

x=2

= ln u x

=1

= ln x3 - x2 + 5x - 3

2 1

= ln (2)3 - (2)2 + 5(2)- 3 - ln (1)3- (1)2+ 5(1) - 3 = ln11 - ln2 2 2 3x - 2x 5+ dx » 1.70 ò1 x3- x2 + 5x - 3

5-29 ..

Calculus

Chapter 5

13x 6x2+1dx 7. 2ò

We will solve the problem using the substitution method of integration. Let u=6x2-1,therefore du=12xdxÞdu=4(3x)dx.Rewrite the original integral in terms of the variable u and solve.

1 x=2 1 2 2 ò13x 6 x +1 dx = òx=1u 2 du 41 x= 2 4 2 3 =1× u 2 3 x=1 2 6 3 +1) (6x2 2 =1 1

6

3 1 = (6(2)2 +1) 2 - 6 (6(1)2 +1) 2

=

3

125 6

343 6

2

2

ò1 3x 6 x +1 dx » 17.75

In Problems 8 and 9, find the average value of the function over the given interval.

8.

f(x)=8x3-16x+4; [1,4] 1

1

b

4

3

b-aò f (x) dx= 4 - 1 ò1 (8x - 16 x + 4) dx 1 =3

x ) (2x4-8x2+4 4

1

1

((2(4)4 - +8(4)2 4(4))- (2(1)4- 8(1)2+ 4(1))) 3 (400- (-2)) =1 3 =134 =

5-30 ..

Calculus

Chapter 5

9. g(x)=2e-0.3x

[0,10]

We will solve the problem using the substitution method of integration. Let u=-0.3x,therefore du=-0.3dx.Rewrite the original integral in terms of the variable u and solve

1 b 1 10 - 0.3x ) dx òaf (x)dx = ò (2e b- a 10 - 0 0 2 x=10 u (e ) du ò=0 (10)(-0.3) x 2 æ x=10ö = - çèeu x ÷ 3 =0ø - 2 æe- 0.3x10ö ç 3è ÷0ø 2 e =0.3(10) --0.3(0) 3 2 =3 » 0.63

(e-

)

(e-31)

10. The total cost (in dollars) of manufacturing x units of a product is C(x)=30,000+250x. a. Find the average cost per unit if 400 units are produced. [Hint: Recall that C(x) is the average cost per unit.]

C(x) x 30, 000 +250x C(x) = x 30, 000 C(x) = + 250 x C(x) =

C(

30, 000 = 400 + 250

C400) = 75 + 250 C

(

= 325

400)

Therefore, the average cost for 400( units is $325 per unit.

400)

5-31 ..

Calculus

Chapter 5

b. Find the average value of the cost over the interval [0, 400].

1 b- a

1 8 b òaf (x)dx = 400- 0ò2(30,000+250x)dx = =

1 æç30,000x+125x2400ö è 0 ÷ 400 ø 1

(((30,000)(400)+125(400)2)) 400 (32, 000, 000 - 0) ((30, 000)(0)+125(0)2)

= 1 400 = 80, 000 Therefore, the average value of the cost will be $80,000.

11. A company manufactures a product and the research department produced the marginal cost function

x C'(x)=300-

0 £ x £ 800

4

where C'(x)is in dollars and x is the number of units produced per month. Compute the increase in cost going from a production level of 400 units per month to 800 units per month. Set up a definite integral and evaluate it.

x b 800 òaC '(x)dx= ò400 (300 - ) dx 4 800

æ x2ö = ç300x ÷ è 8 ø

400

æ (800)2 ö æ ( 400) 2ö ÷-ç 300(400)= ç300(800) 8 ÷ è 8 ø è ø =160,000 - 100,000 b = 60, 000 òaC'(x)dx

Therefore, the increase in cost going from 400 units per month to 800 units per month is $60,000.

5-32 ..

Calculus

Chapter 6

Name ________________________________ Date ______________ Class ____________

Section 6-1 Area Between Curves Goal: To find the area between two curves

Theorem :

Area Between Two Curves

If f and g are continuous and f(x)³ g(x)over the interval [,a b], then the area bounded by y=f(x)and y=g(x)for a£x£bis given exactly by

ab[f(x)-g(x)]dx A=ò Definition:

Gini Index of Income Concentration

If y=f(x)is the equation of a Lorenz curve, then

Giniindex=2ò 1 f(x)]dx 0[x-

In Problems 1–7, find the area bounded by the graphs of the indicated equations over the given interval (when stated). Compute answers to three decimals, if necessary.

1. y=4x3+5;y=0;1£x£5

A = ò15 [4 x3 + 5 - 0] dx 3+ 5 A = ò1 [4 x 5] dx A =x4+ 5 x

A=

5 1

+((1)4 5(1)) 4 ((5)+5(5))-

A = (625 + 25) - (1+ 5) A= 644

6–1 ..

Calculus

Chapter 6

2. y=-2x3+9;y=0;

-3£ x£ 2

A = ò-23[ - x3 + 9 - 0] dx 3+ 2 A = ò- 32 x 9] dx 2 [x4 A= + 9x 2

-3

æ (2)4 ö æ (- 3)4 ö A =ç+9(2)÷ - ç - + 9(- 3)÷ 2 è 2 ø è ø 81 A = (-8 +18)- ( - - 27) 2 A =10 - ( 67.5) A = 77.5

3. y = 5e x - 3; y = 0; 0 £ x £ 2 2 ex A = ò05 [ -3-0]

dx

A = ò02 [5 ex - 3] dx A = 5ex - 3x

A=

2 0

- 3(0)) (5e2 - 3(2))(5e0

A = (36.9452805 - 6) - (5) A » 25.945

6–2 ..

Calculus

Chapter 6

4. y=3x+7;y=-4;-7£x£0 0 A = ò-7[3x+7 - (- 4)] dx 0 A = ò-7[3x+11]dx

3x2 A = 2 +11x 3(0)2

æ A=ç è 2

0

-7

ö æ3(- 7)2 ö +11(0)÷- ç +11(-7) ÷ 2 øè ø

147 A = (0 - 0)- ( - 77) 2 A = 0 - ( 3.5) A = 3.5 5. y=x2-3;y=6 The first step in the problem is to find where the two functions intersect. Set both equations equal to each other to find the intersection points.

x2-3 = 6

x2 = 9 x = 3 ± A graph of y=x2-3and y=6shows that 6³ find the area using the definite integral from –3 to 3.

3 - (x2 - 3)]dx A = ò-3[6 3 A = ò-3[ - x 2 + 9]dx

x3 A = - 3 + 9x

3

-3

ö æ (- 3)3 æ (3)3 A =ç+ 9(3)÷- ç3 è 3 ø è

ö + 9(- 3) ÷ ø

A = (-9 + 27) - (9- 27) A = 18 - (-18) A = 36 6–3 ..

x2 - 3on the interval [–3, 3]. Now

Calculus

Chapter 6

6. y=x2-2;y=x-5;-3£x£2

A = ò-23[( x2 - 2)- (x -5)]dx A = ò-23[ x2 - x+ 3]dx x3 x2 A= + 3x 3 2

2

-3

ö æ(- 3)3 (- 3)2 æ(2)3 (2)2 ö A=ç + 3(- 3)÷ + 3(2) ÷ 3 2 2 è 3 çø ø è ö æ 9 æ8 ö 9- - 9 A =èç - 2 + 6 3 ÷ -÷ ç-è 2 ø ø A = 6.6666667 - (- 22.5) A » 29.167

7. y = 2 x + 3; y = 2 x - 4; - 5 £ x £ 3 3 (2x - 4)] dx A = ò-5[(2x+3)3

A = ò-5[7]dx A = 7x 3

-5

A = 7(3) - 7(-5) A = 21 + 35 A = 56

6–4 ..

Calculus

Chapter 6

In Problems 8 and 9, use a graphing calculator to graph the equations and find relevant intersection points. Then find the area bounded by the curves. Compute answers to three decimal places.

8. y=x3; y = 9 x

Based on the graph above, the points of intersection are –3, 0, and 3, therefore the area between0 the curves is

A = ò-3[(x3 )- (9 x)] dx + ò03[(9x) - (x3)] dx 3 x x3]dx 0 A = ò-3[ x3 - 9x] dx + ò0[9 2

3 0ö æ 4 æ x4 ö ÷ 9x x 9 x2 ÷+ç A=ç - 4 ÷ 2 çè 4 ÷ç 2 0ø -3øè

(0)4 9(0)2 ö æ(0)4 9(0)2 ö æ(-3)4 9(-3)2 ö æ A=ç ÷-ç ÷ +4 ç - 2 æ è 4 2 øè 4 2 ø è ÷ æ81 81ö 81 81 æ ö A = (0- 0)- ç÷+(0- 0) - ç - 2÷ ç è4 è4 2ø ø ø è A = 0 - (- 20.25) + 0 - (- 20.25) A = 40.5

6–5 ..

(3)4 4

-

9(3)2 ö 2 ÷ ø

Calculus

Chapter 6

9. y=x2+2x-3;y=-x2-3x+4;

Based on the graph above, the points of intersection are –3.5 and 1, therefore the area between the curves is 1 [(- x2 - 3x+4)-(2+2x-3)]dx A = ò-3.5 x

A = ò1-3.5 [- 2x2 - 5x+7]dx 1

- 2 x3 5x2 +7x A= 3 2

- 3.5

ö æ- 2(- 3.5)3 5(-3.5)2 æ- 2(1)3 5(1)2 ö A= ç + 7(-3.5) ÷ +7(1)÷- ç è 3 2 3 2 øè ø æ- 2 ö æ85.75 61.25 ö - 24.5 A =èç - 2.5+7÷-èç ø 3 3 2 ÷ ø A = 3.8333333- (-26.5416667) A = 30.375

6–6 ..

Calculus

Chapter 6

Name ________________________________ Date ______________ Class ____________

Section 6-2 Applications in Business and Economics Goal: To solve application problems that involve business and economics

Definitions: 1. Total Income for a Continuous Income Stream If f (t) is the rate of flow of a continuous income stream, then the total income produced during the period from t = a to t = b is b Totalincome=òaf(t)dt

2. Future Value of a Continuous Income Stream

If f(t)is the rate of flow of a continuous stream, 0£ t £ T,and if the income is continuously invested at a rate r, compounded continuously, then the future value FV at the end of T years is given by

FV= ò0T f(t)er(T-t)dt =e rTTò0fte()-rt

dt

3. Consumers’ Surplus If (x,p) is a point on the graph of the price–demand equation p = D(x) for a particular product, then the consumers’ surplus CS at a price level of pis x CS=ò0[D(x)-p]dx

4. Producers’ Surplus If (x,p) is a point on the graph of the price–supply equation p = S( x), then the producers’ surplus PS at a price level of pis x PS=ò 0[p-S(x)]dx

6–7 ..

Calculus

Chapter 6

In Problems 1–3, evaluate each definite integral to two decimals.

4 -5 tdt 1. ò0e

4 - 5t

ò0e

1 dt = - 5 e -5t

4 0

æ 1 ö æ 1 -5(0) ö =ç- e-5(4)÷ ç-è ÷ - 5e è 5 ø ø = 0.20 2. ò07 e0.03(3-2t)dt 0.03(3) 7 - 0.06t 7 dt ò0e0.03(3-2)dt = e ò0 e 7ö æ 1 e-0.06t ÷ =e 0ø ç- 0.06 è 1 æ e-0.06(7)+ 1 e -0.06(0) ö = e0.09 ÷ 0.06 ø èç- 0.06 0.09

» e0.09( - 10.95078+16.66667) » e0.09 (5.71589) » 6.25 40 t) 3. ò0 600e0.04te0.07 ( 40dt 40 t) e 0.07 ( 40dt dt = ò040 600e0.04te2.8-0.07t ò0 600e 0.04t

= ò040 600e2.8- 0.03t dt = 600e2.8ò040 e -0.03t dt 40 ö æ 1 0.03t t) e e0.07(400.04t 600e2.8 eç÷ 600 = dt ò0 è 0.03 0ø 40

æ 1 e-0.03(40) +1 -0.03(0) ö e = 600e2.8 çø è 0.03 0.03 ÷ » 600e2.8 (- 10.03981 + 33.33333) » 600e2.8(23.29352) » 229, 832.29 6–8 ..

Calculus

Chapter 6

4. Find the total income produced by a continuous income stream in the first 6 years if the rate of flow if f(t)=400e0.05t.

Totalincome = òab f ( ) dt 6 te0.05t dt = ò0400

6

400 0.05t = e 0.05 0 400 0.05( 6 ) 400 e 0.05( 0 ) = e 0.05 0.05 » $2799 5. Find the future value, at 1.75% interest, compounded continuously for 7 years, of the continuous income stream with rate of flow f(t)=3000e0.05t.

T F = e rT ò0 f (t) e- rt dt 7 0.05t - 0.0175t V = e(0.0175)(7) ò03000 e dt e 0.0325t F = 3000e0.12257 dt ò0 e

V 3000e0.1225æ0.0325t 7 ö çe FV = 0÷ 0.0325 è ø F 3000e 0.1225 V ( e 0.0325(- 7e)0.0325()0 ) F = 0.0325 V = $26, 654 F V 6. Find the consumers’ surplus at a price level ofp = $160 for the price–demand equation p =D(x) = 460- 0.06 x.

First find x by substituting the valuep = $160 into the price–demand equation.

p = 460 - 0.06x = 460 - 0.06 x 160 =- 0.06 x -300 =x 5000 6–9 ..

Calculus

Chapter 6

Now use the consumers’ surplus definition.

( )- p]dx C = ò0x [Dx [4600.06x- 160]dx S = ò05000 [300- 0.06x]dx C = ò05000 S = 300x- 0.03x2 5000 0 C = (300(5000)- 0.03(5000)2)- (300(0)- 0.03(0) 2 ) S =1,500,000- 750,000- 0- 0 C = $750, 000 S 7. Find C the producers’ surplus at a price level of p=$98for the price–supply equation p=S(x)=10+0.1x+0.0003x2.

S C First find xby substituting the value p=$98into the supply–demand equation.

S C 2

p =10+0.1x+0.0003 x S 2 98 =10+0.1x+0.0003 x

2

0 = - 88 + 0.1x+0.0003 x

x = 400.

Using the quadratic equation will yield that Now use the producers’ surplus definition. x P = ò0[ p-Sx ( )] dx

S = ò0400 [98 - (10 + 0.1x+ 0.0003x2)]dx P = ò0400 [88- 0.1x- 0.0003x2]dx S = 88x- 0.05x2 - 0.0001x3 400 0 P = (88(400)- 0.05(400)2 - 0.0001(400)3)- (88(0) - 0.05(0) 2- 0.0001(0)3) S = (35,200- 8000- 6400)- (0- 0- 0) P = $20, 800 S P S P S P S

6–10 ..

Calculus

Chapter 6

8. Find the consumers’ surplus and the producers’ surplus at the equilibrium price level for the given price–demand and price–supply equations. Include a graph that shows the equilibrium point (x,p). Round all values to the nearest integer if necrssary.

p=D(x)=55-0.3x

p=S(x)=10+0.15x

The equilibrium point is also the point of intersection. Through a process that narrows the window required to see the point of intersection, the graph will look as follows

Therefore, the equilibrium point is (100, 25).

The consumers’ surplus will be

C = ò0x [ D( x)- p]dx S = ò0100 [55- 0.3x- 25] dx C = ò0100 [30- 0.3x]dx S = 30x- 0.15 x2 100 0 C = (30(100) - 0.15(100)2 ) -(30(0) - 0.15(0)2) S = (3000- 1500)- (0- 0) C = $1500

S C S C S C S

6–11 ..

Calculus

Chapter 6

The producers’ surplus will be

P = ò0x [ p-Sx ( )] dx S = ò100 0 [25- (10+0.15x)] dx P = ò100 [15 - 0.15x] dx 0 S =15x- 0.075 x2 100 0 P = (15(100)-0.075(100)2)- (15(0)- 0.075(0)2) S = (1500 - 750) - (0 - 0) P = $750

S P S P S P S

6–12 ..

Calculus

Chapter 6

Name ________________________________ Date ______________ Class ____________

Section 6-3 Integration by Parts Goal: To solve integration problems using the integration by parts formula

Formula:

Integration by Parts òudv=uv-òvdu

Summary :

Integration by Parts: Selection of u and dv

For òudv=uv-òvdu, 1. The product u dv must equal the original integrand.

2. It must be possible to integrate dv (preferably by using standard formulas or simple substitution) 3. The new integral òvdushould not be more complicated than the originaludv. ò 4. For integrals involving xpeax,try u=xp and dv=eaxdx 5. For integrals involving xp(lnx)q,try

u=(lnx)q

and

dv=xpdx

6–13 ..

Calculus

Chapter 6

In Problems 1 and 2, integrate by parts. Assume that x > 0 whenever the natural logarithm function is involved.

1.

òxe7x dx

Let u=xand dv=e7xdx, then du=dxand

v = 17 e7 x. Substituting these terms into

the integration by parts formula yields

òudv = uv - òvdu 1 e7x dx òxe7x = x(1 e7x)-ò7 7

dx ò xe7x ò dx

=x 7

e7x7 7 - 1(1 e7x)+C

1 = x e7x - e7x +C 49 7

xe7x dx 2.

3òxlnxdx Let u=lnxand dv=x3dx,then du = 1x dx and v = 1 x4 . Substituting these terms into 4

the integration by parts formula yields

vdu òudv = uv - ò 4 4 3 òx l xdx = lnx ( 1x4 )- ò41 x (x1 ) dx ln 3 4 ln - 1 x3 dx x òx n xdx = 1 xx ò 4

4

3l

òx

xdx = 1

4

x -1 4

4

x C (14)+ 4

1 4 1 4 n x +C òx3ln xdx = x ln x 16 4 l n

6–14 ..

Calculus

Chapter 6

Problems 3–8 are mixed – some may require use of the integration-by-parts formula along with techniques learned earlier. Assume that g(x)>0whenever lng(x)is involved.

5ò 1ln3xdx

3.

Use integration by parts by letting u=ln3xand dv=dx,then du = 1x dx and v = x.

Substitute these terms into the integration by parts formula yields 5 1 15- ò1x( )dx 5 5 1- ò11dx

5 (ln3x) ò1ln3xdx = (x) (ln3x) (x) =

5

= xln3x-x 1

= (5ln 3(5) - (5)) - (1ln 3(1) -(1)) = 8.540251- 0.098612 5

ò1 ln3xdx » 8.442

ò9xexdx

4.

Use integration by parts by letting u=9xanddv = exdx, then du = 9dx and v = e x .

Substituting these terms into the integration by parts formula yields

= 9 x( ex ) - 9 òe x dx

ò9 xex ò9 dx

= 9 xex -9e x +C

xex dx 3 3x5ex dx ò

5.

3

Use integration by parts by letting u = x3 and dv = 3x2ex dx , then du = 3x2 dxand 3

v = e x . Substituting these terms into the integration by parts formula yields 3

3

x3 x x ò3x5 dx = x3e( ) - òe (3x2)dx x3 dx = x3ex3 -ex 3 +C ò3e

x5 e 6–15 ..

Calculus

Chapter 6

4e4x +3 dx ò e4x +3x Use substitution by letting u=e4x+3xand du=4e4x+3.Substitute these terms into the original equation yields 6.

44x 1 e +3 ò 4x x dx = ò du e +3 u = ln u + C

= ln e4x +3x+ C 7

.

0xe2x2 2 dx ò

.

Use substitution by lettingu = - 2 x2 and du = - 4 xSubstitute these terms into the original equation yields -2x2

2 ò0xe

1 x=2 u òx=0 e du 4 1 4 x=2 1 = - 4 eu 1 x=0 2 4 e2x2 =-

dx = -

0

=

e-2(2)21+ - 2(0)2 4 e

-

0.00008387 + 0.25 »» 0.2499 8.

ò(x+7)8(x-3)dx Use integration by parts by letting u=x-3and dv=(x+7)8dx, then du= dx and

v = 19 (x+7)9. Substituting these terms into the integration by parts formula yields 9 1 ò( x + 7)8 ( x- 3) dx = ( x - 3)( 91 (x + 7)9) - ò9 (x+7) dx 9

= 1 (( x - 3)( x + 7) - 1 ò(x+7) dx 9 9( 9 - 1 1 (x+7)10 = 1 x - 3)( x +

(

9

7) ò( x + 7)8 ( x- 3) dx = 1 x - 3)( x + 9

9

6–16 ..

7) 9

)+C

9 10 - 1 (x+7)10 + C 90

Calculus

Chapter 6

Name ________________________________ Date ______________ Class ____________

Section 6-4 Other Integration Methods Goal: To use the Trapezoidal Rule and Simpson’s Rule to approximate definite integrals.

Use integration tables to solve indefinite integrals

Trapezoidal Rule: Let f be a function defined on an interval [a, b]. Partition [a, b] into n subintervals of equal length Dx=(b-a)/n with endpoints a=x0<x1<x2<L<xn=b Then Tn=[f(x0)+2f(x1)+2f(x2)+L+f(xn)]Dx/2 b

is an approximation of òf(x)dx. a

Simpson’s Rule: Let f be a function defined on an interval [a, b]. Partition [a, b] into n2 subintervals of equal length Dx=(b-a)/n with endpoints a=x0<x1<x2<L<xn=b Then S2n=[f(x0)+4f(x1)+2f(x2)+4f(x1)+L+f(xn)]Dx/3 b

is an approximation of òf(x)dx. a

Process:

Using the integration tables

1. Determine which category has the form of the given integrand (integrals involving trigonometric functions, integrals involving lnu,etc.)

2. Examine the integrals within the category to see if one represents exactly the integrand in the given problem. a. If an exact match is found, use the formula to find the indefinite integral. b. If an exact match is not found, use substitution to change the integral into one that appears in the table. Use the formula to find the indefinite integral.

Replace the original function after the formula is used.

6–17 ..

Calculus

Chapter 6

Use Table II (Integration Table) to find each integral in Problems 1–5.

x2

1. ò

16 x2 - 25

dx

Examining the form of the integral, we should look at formulas 40–45, which involve

u2-a2. Based on this, we let u2 =16x2 and a 2 = 25 so that 16x2-25=

u2-a2.

Then we have u=4x,x2=1u2,and du=4dx. Using formula 44 and the substitutions, the 16

integral will be written

ò

x2 2

16 x - 25

1 dx = 4 ò =

1

1 2 16 u 2

u -52

du

u2

du u 2 - 52 u 1æ1 = ç u 2 - 52 + 52lnu+ 64è2 =

64

ò

u2 - 52 ö C ÷+ ø

1æ1 ö x2 x2 ç 4 x 16 - 25 + 25ln 4x + 16 - 25 ÷+C 64è2 ø

1 2. 3ò dx 1 x(7+3x) Examining the form of the integral, we should look at formulas 3–12, which involve a+bu. Based on this, we let u = x. Therefore, we use formula 9 with a=7and b=3. Use this information to rewrite and solve the integral as follows: 3

ò1

1 1 x=3 dx = òx=1 u(a+bu) du x(7+3x) x=3

1 u = ln a + b u x a 7 + 3 x x=1 1 3

= ln 7 1 1 3 1 1 - ln = ln 7 7+3(3) 7 7+3(1) » 0.0898 6–18 ..

Calculus

3. ò

Chapter 6

16 x2 + 36 16 x2

dx

Examining the form of the integral, we should look at formulas 32–39, which involve

u2 + a2. Based on this, we let u2=16x2and 16x2+36= u2+a2. Then we have u = 4 x and du = 4dx. Using formula 35 and the substitutions, the integral can be written as

ò

16 x2 + 36 16 x2

1 u2 +62 dx = ò du u2 4 1æ = ç 4ç è 1 = æ 4ç ç è

ö u2 +62 + ln u + u2 + 62 ÷ u ÷ ø 16x2 +36 4x

ö + ln4x + 16x2 + 36 ÷ ÷ ø

1 3 +5 x 4. ò0 x dx 4 +7

Examining the form of the integral, we should look at formulas 15–20, which involve a + bu and c + du. Based on this, we let u = x. Therefore, we use formula 20 with d = 7. Use this information to rewrite and solve the integral as a=3,b=5,c=4,and follows:

x=1a+bu 1 3 +5 x ò04 + 7 x dx = òx=0 c + du du x=1

=

bu ad - bc ln c + du + d2 d x=0 1

5 x (3)(7)- (5)(4) = + ln 4+7x (7)2 7 0 5x 1 1 = + 49 l n 4 + 7 x 7 0 5(1) 1 5(0) 1 + ln 4+7(1) + 49 ln4 + 7(0) 7 49 7 » 0.7915 =

6–19 ..

Calculus

Chapter 6

1 5. ò dx 49- 16x2 Examining the form of the integral, we should look at formulas 13 and 14 which involve a2-u2. Based on this, we let u2=16x2and 49-16x2=a2-u2. Then we have a = 7, u = 4x, and du=4dx. Using formula 14 and the substitutions, the integral can be written as

1 1 1 du dx= 4 ò ò 7 2 - u2 49 - 16 x2 u +a ö 1æ1 = ln u -a÷ 4çè2a 4 x +ø7 ö = 1 æ 1 ln -7÷ø 4çè2(7) 4 x 1 4x+7 = ln 56 4x- 7

8

6. Use the trapezoidal rule with n = 4 to approximate ò1+x3dx.

0

Partition [0, 8] into 4 equal subintervals of width (8 – 0) / 4 = 2. The endpoints are x0=0,x1=2,x2=4,x3=6, and x4=8. We calculate the value of the function at each endpoint: x 0 2 4 6 8

f(x) 1 3 8.0623 14.7309 22.6495

By the trapezoidal rule:

(6)f (8)]+f (4)+2 Tn= [ f(0)+2 (2)+2 f f Dx/2 = [1 + 3 + 8.0623 +14.7309 + 22.6495](1) = 49.44

6–20 ..

Calculus

7.

Chapter 6

7 1 Use Simpson’s rule with n = 3 to approximate ò dx.

11+ x2

Partition [1, 7] into 6 equal subintervals of width (7 – 1) / 6 = 1. The endpoints are x0=1,x1=2,x2=3,x3=4,x4=5,x5=6, and x6=7. We calculate the value of the function at each endpoint: x 1 2 3 4 5 6 7

f(x) .5 .2 .1 .0588 .0385 .0270 .02

By the simpson’s rule:

+ f (5) + 4 f(6) +f(7)]Dx/3 (3)+4 (2)+2 S6 = [ (1)+4 f f f (4) 2f = [.5 + .2 + .1 + .0588 + .0385 + .0270 +.02](1 / 3) = .31

6–21 ..

Calculus

Chapter 6

6–22 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-1 Functions of Several Variables Goal: To work with functions that involve more than one variable

Definition:

An equation of the form z=f(x,y)describes a function of two independent variables if, for each permissible ordered pair (x, y), there is one and only one value of z determined by f(x,y).

Examples of such functions are:

A(x,y)=xy

Area of a rectangle

V(r,h)=r2h

Volume of a right circular cylinder Compound Interest

nt

æ rö A(P,r,t,n)=P1 ç + ÷nø è

1.

If f (x,y) = 4x3 + 2 y2 + 5 x +15, find f(2,3). f ( x , y) = 4 x2 + 2y2 +5x+15

f (2, 3)= 4(2) 2 + 2(3)2 +5(2)+15 f (2, 3)= 4(4) + 2(9)+10+15 f (2, 3)= 59 2.

8 x +5 y ,find g (0,2).

7 x +10 y 8x+5y g ( x , y) = 7x+10y 8(0) + 5(2) g (0, 2) = 7(0)+10(2)

If g(x,y) =

0+10 0+20 g (0, 2) = 0.5

g (0, 2) =

7–1 ..

Calculus 3.

Chapter 7

If f(x,y,z)=5x3+2y2-4z3+5,find

f (1, 2, 3).

f ( x , y, z) =5 x + 32 y 2 - 4 z 3 +5 f (1, 2, 3)= 5(1)3 + 2(2) 2 - 4(3) 3 +5 f (1, 2, 3)= 5(1) + 2(4)- 4(27)+5 f (1, 2, 3)= 5 +8 -108+5 f (1, 2, 3)= - 90

4.

Find A(2000,0.03,10) for A(P,r, t) = P + Prt.

A(P,r,t) = P + Prt A(2000,0.03,10)= 2000 + 2000(0.03)(10) A(2000,0.03,10) = 2000 + 600 A(2000,0.03,10) = 2600

5.

Find v(10,6,5) for v(l, w, h)= lwh.

v(l,w,h) = lwh v(10,6,5)= (10)(6)(5) v(10,6,5)= 300

In Problems 6 and 7, find the indicated function f of a single variable.

6.

f ( y) = G(3, y)for G(x,y)=x3+2xy2-4y+3. f ( y) = G(x,y) = x3 + 2xy2 - 4y + 3 f ( y) = G(2,y) = (3)3 + 2(3)y2 - 4y + 3 f ( y) = G(2,y) = 27 + 6 y2 - 4 y + 3 f ( y) = G(2,y) = 6y 2 - 4y + 3 0

7–2 ..

Calculus 7.

Chapter 7

f(x)=N(x,3x)for N(x,y)=x2+y2+7xy-2x-7.

f ( x) = N(x,y) = x 2 + y 27+ x y2 -x - 7 + x) 2 7+ (3xx)- 2x- 7 f ( x) = N(x,3x = x 2 (3 = x2 + 9x2 +21x2 - 2 x - 7

f ( x) = )

f ( x) = N(x,3x = 31x2 - 2 x- 7 ) N(x,3x 8.

) The Cobb-Douglas production function for a company is given by

f(x,y)=40x0.2y0.8 where x is the utilization of labor and y is the utilization of capital. If the company uses 1500 units of labor and 1200 units of capital, how many units will be produced by the company?

f ( x, y ) = 40 x0.2 y0.8 , 1 2 0 0 ) f ( 1 5 0 0= 40(1500)0.2(1200)0.8 » 40( 4.3174)( 290.6330) f (1500,1200) » 50,191 f (1500,1200)

7–3 ..

Calculus

Chapter 7

7–4 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-2 Partial Derivatives Goal: To find partial derivatives of functions of several variables

Definition:

Partial Derivative

With Respect to x:

With Respect to y: Definition:

¶z f (x,y + h)- f (x,y) ¶x = fx (x,y) = lim h h®0 ¶z ¶y f (x,y + h)- f (x,y ) = fy (x,y) = lim k k®0

Second–Order Partial Derivatives

If z = f(x,y),then fxx= fxx(x,y) = fxy= fxy(x,y) =

fyx= fyx(x,y) =

¶2z ¶x2 ¶2z

=

¶ æ¶z ö ç ÷ ¶xè

¶ æ¶z ö = ¶yèç ÷ ¶y¶x ¶ æ¶zö

¶ 2z = ç ÷ ¶x¶y ¶xè¶yø

¶ 2 z ¶ æ¶zö fyy= fyy(x,y) = ¶ = ç ÷ y2 ¶y è¶y ø

7–5 ..

Calculus

Chapter 7

In Problems 1–3, find the indicated first-order partial derivative for each function z=f(x,y).

1.

f y(x,y)if f(x,y)=9x3y+5xy4-4y+8x-12.

We are looking for the partial derivative with respect to y. Therefore, we consider only the y as a variable and all other variables as constants.

f ( x,y ) = 9 x3y+ 5x 4 - 4 y+8x-12 f y ( x,y ) = 9 x3 + 20y 3 - 4 x y 2.

fx(x,y)if f(x,y)=7xy-2x+8y-11xy3+17.

We are looking for the partial derivative with respect x. toTherefore, we consider only the x as a variable and all other variables as constants.

f ( x,y ) = 7 xy - 2 x + 8y -11xy3+17 fx ( x,y ) = 7 y - 2-11 y3

3.

¶z if z=2y2-2xy+7x-5. ¶x

We are looking for the partial derivative with respect x. toTherefore, we consider only the x as a variable and all other variables as constants.

z = 2 y2 - 2 xy+7x-5 ¶z = - 2 y +7 ¶x

7–6 ..

Calculus 4.

Chapter 7

Find the value of fx(0,3)if f (x,y) = 3x3 y2 +5xy- 8 x + 2 y +15.

We will first find the partial derivative with respect to x. Therefore, we consider only the x as a variable and all other variables as constants.

f ( x,y ) = 3x3 y2 + 5xy - 8 x + 2 y + 1 5 fx( x,y ) = 9x2 y2 +5 y - 8 Now find the value using the partial derivative.

fx ( x , y) = 9 x 2y 2 +y 5- 8 fx (0, 3)= 9(0)2(3)2 +5(3)- 8 fx (0, 3)=15-8 fx (0, 3)= 7

5.

Find the value of f y (3,1)if f (x,y) = 5 x2 - 7x2y+ 2xy- 8 y + 6.

We will first find the partial derivative with respect to y. Therefore, we consider only the y as a variable and all other variables as constants.

f ( x,y ) = 5 x2 - 7 x2 y + 2 xy-8y+6 f y( x,y ) = - 7 x2 + 2 x - 8 Now find the value using the partial derivative.

f y ( x , y) = 7 x 2 +x -2 8 f y (3,1) -

7(3)2 + 2(3) - 8

f y (x,y) = 6 3 + 6 - 8 f y (x,y) - 65 = = 7–7 ..

Calculus

Chapter 7

In Problems 6–8, find the indicated second-order partial derivative for each function f(x,y).

6.

fxx(x,y)if f(x,y)=7x2+2y2-8xy+17

We will first find the partial derivative with respect to x. Therefore, we consider only the x as a variable and all other variables as constants.

f (x, y) = 7x2 +2y2 - 8xy+17 fx(x,y)=14x- 8y Now using the function just found, find the partial derivative with respect to x.

fx(x,y)=14x-8y fxx(x,y)=14 7.

3222 fxy (x,y)if f(x,y)=ex+y+6xy

We will first find the partial derivative with respect to x. Therefore, we consider only the x as a variable and all other variables as constants.

f ( x,y ) = e

3x2+2y2+6 xy 2

2

fx( x,y ) = (6x + 6 y)(e3 x +2 y +6xy ) y.to Now using the function just found, find the partial derivative with respect 3x2+2y2+6 xy

fx ( x,y ) = (6 x+6y)(e

)

2

2

2

2

x e3 x +2y +6 xy) +(e3x +2 y +6xy )(6) fxy ( x,y ) = (6 x+6y)(4y+ 6)( 2

fxy ( x,y ) = [( x+6y)(4y+ 6 x) + 6](e3 x +2

y2+6xy

)

32+2 y2+6xy

fxy ( x,y ) = 6 (24 xy+36x2 + 242y + 36 xy + 6)(e x 2

2

fxy ( x,y ) = (36x2 +24y2+ 60 xy + 6)(e3x +2 y +6xy )

7–8 ..

)

Calculus 8.

Chapter 7

f yx(x,y)if f(x,y)=(7x+2)7

We will first find the partial derivative with respect to y. Therefore, we consider only the y as a variable and all other variables as constants.

f ( x,y ) = (7x+2y)7 f ( x,y ) = 7(7x+2y)6(2) y ( x,y ) =14(7x+2y)6 f Now using the function just found, find the partial derivative with respect x.to y f y ( x,y ) =14(7 x + 2y)6

f yx( x,y ) = 84(7 x + 2y)5(7) f yx( x,y ) = 588(7 x + 2y)5

In Problems 9–12, find the indicated function or value if N(x,y)=12x3+5xy-7y2+4x-9y+17and S(x,y)=x2 ln y + 3ye2 2 x.

9.

Nyy(x,y) N(x,y) =12x3 +5xy - 7 y2 + 4 x - 9 y +17 Ny(x,y) = 5x- 14 y - 9 Nyy(x,y)= 14 -

10.

S y(4,-4) S(x,y) = x2lny+3y2e2 x x2 Sy (x,y) = +6ye2x y (4)2 S y (4 - 4) = + 6(- 4)e2(4) 4 » S y , - 4) 4 - 71,542.9917 S y (4 - 4) 71,546.9917 » , (4 ,

7–9 ..

Calculus 11.

Chapter 7 Sx(x,y)

S(x,y) = x 2 ln y + 3 y2e 2 x 2 2x Sx(x,y) = 2xln y + 6 y e

12.

Nxx(7,3) N(x,y) =12x3 +5xy - 7 y2 + 4 x - 9 y +17 Nx(x,y) = 36x2 +5y + 4 Nxx(x,y)= 72x = 72(7) N xx (7, 3) = 504 N xx (7, 3)

13.

A firm produces two types of toy trains each week,ofx type A and y of type B. The weekly revenue and cost functions (in dollars) are

R(x,y)=50x+70y+0.07xy-0.04x2-0.04y2 C(x,y)=7x+9y+18,000

Find the profit function and then find P x(800,1000)and Py(800,1000), and interpret the results.

P(x,y) = R(x,y)- C(x,y) P(x,y) = 50x+70 y+0.07xy - 0.04 x2 - 0.04 y2 - (7x+9y+ 18, 000) P(x,y) = 43x+61 y+0.07xy - 0.04 x2 - 0.04 y2 - 18, 000 P(x,y) = 43x + 61y+0.07xy- 0.04x2 - 0.04 y2 -18,000 Px(x,y) = 43 + 0.07y- 0.08x P (800,1000) = 43 + 0.07(1000) - 0.08(800) x (800,1000) = 43 + 70- 64 P (800,1000) = 49 x P x

P(x,y) = 43x+61y+0.07xy- 0.04x2 - 0.04 y2 -18,000 61+0.07x- 0.08y Py(x,y) = P (800,1000) = 61+ 0.07(800) - 0.08(1000)

y (800,1000) = 61+56- 80 P (800,1000) = 37 y P y

7–10 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-3 Maxima and Minima Goal: To find local maxima and minima of functions of several variables

Theorem 1 :

Local Extrema and Partial Derivatives

Let f(a,b)be a local extremum (a local maximum or a local minimum) for the function f. If both fx and f y exist at (a, b), then fx (,a b)=0and f y (,a b)=0.

Theorem 2: Second-Derivative Test for Local Extremum If 1.

2.

3.

4.

z=f(x,y) fx (,a b)=0and f y (a,b) = 0 [(a, b) is a critical point] All second–order partial derivatives of f exist in some circular region containing (a, b) as center.

A = fxx(a,b), B = fxy(a,b), C = fyy(a,b) Then Case 1: If A - B2> 0 and A < 0, then f(a,b)is a local maximum.

Case 2: If C - B2> 0 and A > 0, then f(a,b)is a local minimum.

Case 3: If AAC-B2 < 0, then f has a saddle point at (a, b).

C - B2=0,the test fails.

Case 4: If AC

1.

If f(x,y)=17x-12y+27,find

fx (x,y) and f y (x,y ), and explain, using

Theorem 1, why f(x,y)has no local extremum.

f(x,y)=17x-12y+27 fx(x, y) =17

f(x,y)=17x-12y+27 fy(x, y) = - 12

Since the partial derivative with respect to x and y cannot equal zero, there is no local extremum.

7–11 ..

Calculus

Chapter 7

In Problems 2–5, use theorem 2 to find the local extremum. f(x,y)=5x2-3y2+2x-3y+17 2.

f ( x, y) = 5x2 - 3 y2+2x-3y+17 fx( x, y) =10 + 2 0 =x + 2

f ( x, y) = 5x2 - 3 y2+2x-3y+17 f y( x, y) =- 6 y - 3 0 = 6y- 3 3 - 6y

- 2 =10 - 51 =xx - 21 = y 10 x partial derivatives, there is a critical value at the point (- 1 , - 1 ).

Based on the above 5

2

Now find the values for A, B, and C.

B=f xy(a,b)

A=fxx(a,b)

C=fyy(a,b) C =-6

B=0

A=10

(10)(-6) -(0)2 AC - B2=AC =- 60 - B2 - 15 , - 12 ).

Since the above value is negative, the function f has a saddle point(at 3.

f (x,y) = 2 x2 + 4 y2 + 7 x - 3 y -12 f ( x, y) = 2x2 + 4 y2 + 7 x - 3 y -12 fx( x, y) = 4 + 7 0 = x +7

f ( x, y) = 2x2 + 4 y2 + 7 x - 3 y -12 f y( x, y) = 8y - 3 0 = 8y - 3 3 = 8y

- 7 =4 3= - 47 = x 8 y 4 (- 52 ,6 1).

Based on the above partial derivatives, there is a critical value at the point x Now find the values for A, B, and C. x B=f xy(a,b) A=fxx(a,b) C=fyy(a,b) C =8 A=4 B=0 AC-B2=(4)(8)-(0)2 AC-B2=32 Since the above value is positive and the value of A is positive, the function has a minimum at the point(- 74 ,8 3).

7–12 ..

Calculus 4.

Chapter 7

f(x,y)=4xy+5y2-4x-7y+19 f ( x, y) = 4xy + 5 y2-4x-7y+19 fx( x, y) = 4 - 4 0 =y - 4

f ( x, y) = 4xy + 5y2 -4 x-7y+19 f y( x, y) = 4x +10y- 7 0 = 4x +10(1) - 7 - 3 = 4x

4 =4 1=y - 43 = x 4 y partial derivatives, there is a critical value at the point Based on the above (- 3 ,1).

4

Now find the values for A, B, and C.

B=f xy(a,b)

A=fxx(a,b)

C=fyy(a,b) C =10

B=4

A=0

AC-B2=(0)(10)-(4)2 AC-B2=-16 Since the above value is negative, the function f has a saddle point(at - 3 ,1).

4

7–13 ..

Calculus 5.

Chapter 7

f(x,y)=1x4+y2-4xy+3 2

f ( x,y ) = 21 x4 + y2 -4xy+3

f ( x,y ) = 12 x4 + y2-4xy+3 f y ( x,y ) = 2 y 4 x

fx ( x,y ) = 2 x3 - 4 y

Setting both equations equal to zero and solving will yield the critical points of (0, 0), (2, 4) and (–2, –4). Now find the equations for A, B, and C. A=fxx(a,b)

B=f xy(a,b)

A = 6x2

B=-4

C=fyy(a,b) C = 2

For the critical point (0, 0), the value of A will be 0. Therefore,

AC-B2=(0)(2)-(-4)2 AC-B2=-16 Since the above value is negative, the function has a saddle point at the point (0, 0).

For the critical point (2, 4), the value of A will be 24. Therefore,

AC-B2=(24)(2)-(-4)2 AC-B2=32 Since the above value is positive and the value of A is positive, the function has a minimum at the point (2, 4).

For the critical point (–2, –4), the value of A will be 24. Therefore,

AC-B2=(24)(2)-(-4)2 AC-B2=32 Since the above value is positive and the value of A is positive, the function has a minimum at the point (–2, –4).

7–14 ..

Calculus 6.

Chapter 7 A company produces two types of toy cars per day: x units of type A and y units of type B. If the revenue and cost equations for the day are (in thousands of dollars)

R(x,y)=20x+25y C(x,y)=2x2-5xy+4y2-5x+9y+7 determine how many of each type of toy car should be produced per day to maximize profit? What is the maximum profit?

P(x,y) = R(x,y)- C(x,y) P(x,y) = 20x+25y- (2x2 - 5 xy + 4 y2 - 5x +9 y + 7) P(x,y) =- 2x2 +5xy- 4y2 + 25 x +16 y - 7 Px(x,y) = 4 x + 5 y + 2 5 Py(x,y) = 5x- 8y+16 Setting both equations equal to zero and solving will yield the critical point of (40, 27). Now find the values for A, B, and C. A=Pxx(a,b)

B=P xy(a,b) B=5

A=-4

C=Pyy(a,b) C =-8

AC-B2=(-4)(-8)-(5)2 AC-B2=7 Since the above value is positive and the value of A is negative, the function has a maximum at the point (40, 27). Therefore, you should produce 40 type A toy cars and 27 type B toy cars to maximize your profit. The maximum profit will be 2 5 1 6 x+ P(x,y) = 2 x2 + 5 -xy4 + y2

y- 7

P(40,27) - 2(40)2+5(40)(27)- 4(27) 2 + 25(40) +16(27)- 7 P(40,27) = $706 -

7–15 ..

Calculus 7.

Chapter 7 A rectangular box with no top and two parallel partitions must hold a volume of 125 cubic inches. Find the dimensions that will require the least amount of material.

Based on the information in the problem, we will let x = length, y = width, and z = height. Therefore, M=xy+2xz+4yzand the volume is V=xyz=125. Solve the volume equation for z and reduce the original equation down to two variables yields

M(x,y)=xy+250+500 y

x

Use the new equation for M and follow the procedure outlined in theorem 2.

M(x,y) = xy +250 +500 x

M(x,y) = xy + 250 +500 x

Mx (x,y) = y - y 2

250y My(x, y)= x - y2

x 500

Setting both equations equal to zero and solving will yield the critical point of (10, 5).

Now find the values for A, B, and C.

(,) A=Mxxab

B = Pxy(a,b)

C = Pyy (a,b)

A = 1000

B =1

C = 500 3

x3

y

A=1

C=4

AC-B 2=(1)(4)-(1)2 AC-B

2

=3

Since the above value is positive and the value of A is positive, the function has a minimum at the point (10, 5). Therefore, the length of the box should be 10 inches, the width of the box should be 5 inches, and solving for the variable z will yield a height of 2.5 inches.

7–16 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-4 Maxima and Minima Using Lagrange Multipliers Goal: To find maxima and minima of functions of several variables using Lagrange multipliers Theorem 1 : Method of Lagrange Multipliers for Functions of Two Variables Any local maxima or minima of the function z = f (x,y) subject to the constraint g(x,y) = 0 will be among those points(x0 ,y0) for which ( x0 , y0 ,is 0 ) a solution of the system Fx(x,y ,) = 0 Fy(x,y,)=0

F (x,y ,) = 0 where F(x,y,) = f (x,y) +g(x,y ), provided that all partial derivatives exist.

Procedure: Method of Lagrange Multipliers: Key Steps Step 1: Write the problem in the form z = f ( x, y ) Maximize (or minimize) subject to g(xy , )=0 Step 2: Form the function F:

F(x,y,)=f(x,y)+g(x,y) Step 3: Find the critical points of F; that is, solve the system

Fx (x,y, ) = 0 Fy(x,y,)=0 F(x,y, ) = 0 Step 4: If ( x0 , y0 ,is ( x0 , y0 ) will 0 ) the only critical point of F, we assume that always produce the solution to the problems we consider. If F has more than one critical point, we evaluate z = f (x,y) at ( x0 , y0 ) for each critical point ( x0 , y0 ,of 0 ) F. For the problems we consider , we assume that the largest of these values is the maximum value off(x,y ), subject to the constraint g(x,y) = 0 and the smallest is the minimum value off (x,y ), subject to the constraint g (x,y) = 0.

7–17 ..

Calculus

Chapter 7

Theorem 2: Method of Lagrange Multipliers for Functions of Three Variables

Any local maxima or minima of the function w=f(x,y,z)subject to the constraint g(x,y,z)=0will be among the set of points (x0,y0,z0)for which (x0,y0,z0,0)is  a solution of the system Fx(x,y,z,)=0 Fy(x,y,z,)=0 Fz(x,y,z,)=0 F(x,y,z, )=0 

where F(x,y,z,)=f(x,y,z)+g(x,y,z),provided that all partial derivatives exist.

1. Use Theorem 1 to explain why no minima exist.

Minimize:

f(x,y)=8x-5y

subject to:

5x+8y=9

The problem will need to be rewritten in standard form first.

Minimize: subject to:

f(x,y)=8x-5y 5x+8y-9=0

Now find the function F using the definition.

F(x,y l ) = f ( x, y) + l g(x,y) , l ) = (8 x 5 -y )+ l (5x + 8y - 9) F(x,y l ) = 8 x - 5 y + 5xl +8yl - 9 l , Find F(x,ythe three partial derivatives. , Fx ( x,l ) = 8 + 5 l y, l )=-5+8l Fy(x,y, l ) = 5 x + 8 y - 9

Fl (x,y, Therefore, Fx = 0 and Fy = 0have no simultaneous solution.

7–18 ..

Calculus

Chapter 7

In Problems 2–5, use the method of Lagrange multipliers.

2. Maximize:

f(x,y)=9xy

subject to:

2x+2y=6

Step 1:

Rewrite the problem.

f(x,y)=9xy 2x+2y-6=0

Maximize: subject to:

Step 2:

Form the function F.

F(x,y, l ) = f (x,y) + l g(x,y) F(x,y, l ) = (9 xy) + l (2x+ 2y- 6) F(x,y, l ) = 9 xy + 2x l + 2yl - 6l Step 3:

Find the critical points.

Fx ( x,l ) = 9 y + 2 l y, l ) =9x = 0 Fy(x,y, l ) = 2 x + 22ly - 6 = 0 =0 Fl (x,y, Therefore, y = - 2l9 and x = - 2l9 .

2 x + 2y - 6 = 0 -

4l 9

-

y = -(

4l 0 9 -6=

y=

-8l = 6

-2(6.75) ) 9

3 2

x = -( x=

- 2 ( 6.75) ) 9

3 2

9

54 -8l = l = - 6.75 Step 4:

æ3 3 27 ö Since the solution ç , 2 , ÷is the only critical point ofF, the 4 ø è2

3 æ 3ö point ç , 2 ÷will result in the maximum value of the function. è2 ø

7–19 ..

Calculus

Chapter 7

3. Minimize:

f(x,y)=2x2+2y2

subject to:

4x+5y=20

Step 1:

Rewrite the problem.

f(x,y)=2x2+2y2

Minimize: subject to: Step 2:

4x+5y-20=0

Form the function F.

F(x,y ) = f (x ,y)+g(x,y) ,

) = (2 x2 +2y2)+(4x+5y - 20)

F(x,y ) = 2 x2 + 2 y2 + 4x+ 5y - 20 , Step 3: Find the critical points.

F(x,y ,Fx ( x,) = 4 x + 4 = 0 ) = 4 y + 5 = 0 y, Fy(x,y, ) = 4 x + 5y - 20 = 0 F(x,y, Therefore, x = - and y = - 5 .

4

4x + 5y- 20 = 0 4(- ) + 5(- 5 )-20 = 0 4 - 4 - 6.25 = 2 - 10.25 = 0 2080 41

x=-

æ 80 ö æ 5 öæ 80 ö x =-ç÷ y =ç- ÷÷ è 41 ø è 4 øçè 41 ø x=

80 41

= -

Step 4:

y = - 5 4

y=

100 41

80 ö æ80 100 ,Since the solution ç , is the only critical point ofF, the 41 ÷ è 41 41 ø

80 100 æ ö point ç , ÷will result in the minimum value of the function. è 41 41 ø

7–20 ..

Calculus

Chapter 7

4. Minimize:

f(x,y,z)=x2+2y2+z2

subject to: Step 1:

2x-2y+3z=-21 Rewrite the problem.

f(x,y,z)=x2+2y2+z2 Minimize: 2x-2y+3z+21=0 subject to: Step 2: Form the function F. F(x,y,z ) = f ( x, y, z) +g(x,y,z)

) = (x2 +2 y2 + z2)+(2x-2y +3z + 21) F(x,y,z ) = x2+y2 2 + z2 +2x- 2y +3z + 21

,

, Step 3: Find the critical points.

F(x,y,z ) = 2 x + 2 = 0 Fx(x,y,z , ) = 4 y - 2 = 0 , Fy(x,y,z Fz(x,y,z,) = 2 z +3 = 0 ,F(x,y,z,) = 2 x - 2 y +3z + 21 = 0 Therefore, x = - , y = 0.5, and z = - 1.5.

2x- 2y+3z+21= 0 2(- ) - 2(0.5) + 3(- 1.5) + 21 - 2- - 4.5 ==- 21 -7.5 =0 - 21 14 5 y =0.5(2.8) y =1.4 7 y= 5

=

x=- æ14 ö x =-ç è 5 ÷ø 14 x=5

z = 1.5(2.8) z - 4.2 21 5 z= =

æ 14 7 21 14 ö Step 4: Since the solution ç- 5 , 5 , - 5- , 5 ÷is the only critical point ofF, è ø 14 5 æ 7 21 ö the point ç- , 5 , will result in the minimum value of the function. è 5÷ ø 7–21 ..

Calculus

Chapter 7

5. Maximize:

f(x,y,z)=5xyz

subject to:

3x+2y+3z=90

Step 1:

Rewrite the problem.

f(x,y,z)=5xyz 3x+2y+3z-90=0

Maximize: subject to: Step 2:

Form the function F.

F(x,y,z,) = f ( x, y, z) +g (x, y,z) F(x,y,z,) = (5 xyz) + ( 3 x +y2+ 3 z - 90) F(x,y,z,) = 5 xyz+ 3x + 2 y+3z - 90 Step 3:

Find the critical points.

Fx(x,y,z ) = 5 yz + 3 = 0 ) = 5 xz + 2 = 0 , Fy(x,y,z ) = 5 xy+ 3 = 0 ,  (x,y,z, ) = 3x + 2 y +390 F z-

=0

Fz(x,y,z, Simplifying the first two equations will yield the resulty =1.5xand simplifying the z = x. first and third equations will yield the result

3 x + 2 y + 3 z - 90 = 0 3x + 2 ( 1 . 5 x ) + 3 (-x 90 ) =0 3 x + 3 x +3 x = 90 9 x = 90 x =10

Step 4:

y =1.5x y =1.5(10) y =15

z=x z =10

5yz + 3 = 0 5(15)(10)+ 3 = 0  = - 250

Since the solution(10, 15, -10, 250)is the only critical point ofF, the

will result in the maximum value of the function. point (10, 15, 10)

7–22 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-5 Method of Least Squares Goal: To find the equation of a line using the method of least squares

Theorem :

Least Squares Approximation

For a set of n points 1,y1),(x2,y2), (x (xn,yn),the coefficients of the least squares line y=ax+bare the solutions of the system of normal equations n æ n x2ö æ n ö k÷a+çå xk ÷b=å xkyk ø è k=1 ø k=1 çå èk=1

n æn ö xk÷a+nb=å yk èk=1 ø çå k=1

and are given by the formulas

a=

æn ö æn öæn ö n çå xkyk÷-øè çå xk÷øçå yk÷ k=1 è k=1 èk=1 ø n

2

æ n x2ö æ ö n ç å k ÷ çå- xk÷ è k=1 k=1 ø

n æn ö å yk - a ç å x k ÷ èk=1 ø b = k=1 n

7–23 ..

Calculus

Chapter 7

In Problems 1 and 2, find the least squares line. Graph the data and the least squares line.

1.

Totals

x 1 2 3 4 10

y 4 5 4 8 21

xy 4 10 12 32 58

x2 1 4 9 16 30

The normal equations would be30a + 10b = 58 10a + 4b = 21 The solutions for aand b are

æn ö æn öæn ö n ç å xkyk ÷- çå xkøçå yk÷ è k=1 ÷èk=1 ø 4(58) - (10)(21) =1.1 a = k=1 = 2 4(30)- (10)2 æn ö æn ö n ç å x2k÷-çå xk è k=1 ø è k=1 ÷ ø n æn ö å yk - a ç å x k ÷ èk=1 ø 21- (1.1)(10) = 2.5 b = k=1 = 4 n

Therefore, the equation of the least squares line is y = 1.1x+ 2.5.

7–24 ..

Calculus

Chapter 7

2.

Totals

x 1 2 3 4 10

y 3 0 –2 –1 0

xy 3 0 –6 –4 –7

x2 1 4 9 16 30

The normal equations would be30a +10b = -7

10a + 4b = 0 The solutions for aand b are

æn ö æn öæn ö ÷ å yk ÷ n çè å xkyk ÷- k=1xkøèçk k=1 =1 ø= 4(-7) - (10)(0) = - 1.4 a= 2 4(30) - (10)2 æn ö æn ö n ç å x2k÷xk è k=1 ø èçå k=1 ÷ ø n æn ö å yk - a ç å x k ÷ èk=1 ø 0 - (- 1.4)(10) 3.5 = b = k=1 = 4 n

Therefore, the equation of the least squares line is y = - 1.4x + 3.5.

7–25 ..

Calculus

Chapter 7

In Problems 3 and 4, find the least squares line and use it to estimate y for the indicated value of x. Round answers to two decimal places, if necessary.

3.

Totals

x 3 5 7 9 11 35

y 2 8 6 7 9 32

xy 6 40 42 63 99 250

x2 9 25 49 81 121 285

Estimate y when x=7.

The normal equations would be285a + 35b = 250 35a + 5b = 32 The solutions for aand b are

æn ö æn öæn ö n ç å xkyk ÷- çå xkøçå y k ÷ è k=1 ÷èk=1 ø= 5(250) - (35)(32) = 0.65 a = k=1 2 5(285) - (35)2 æn ö æn ö n ç å x2k÷-çå xk è k=1 ø è k=1 ÷ ø n æn ö å yk - a ç å xk÷ èk=1 ø 32 - (0.65)(35) b = k=1 =1.85 = 5 n

y = 0.65 x+ 1.85and the value ofy Therefore, the equation of the least squares line is will be estimated to y=0.65(7)+1.85=6.4.

7–26 ..

Calculus

Chapter 7

4.

Totals

x

y

xy

–6 –3 0 3 6 0

12 9 8 5 1 35

–72 –27 0 15 6 –78

x2 36 9 0 9 36 90

Estimate y when x=9.

The normal equations would be90a +0b=- 78 0a +5b= 35 The solutions for aand b are

æn ö æn öæn ö n ç å xkyk ÷- çå xkøçå yk÷ è k=1 ÷èk=1 ø 5(- 78) - (0)(35) = - 0.87 a = k=1 = 2 5(90) - (0)2 æn ö æn ö n ç å x2k÷-çå xk è k=1 ø è k=1 ÷ ø n æn ö å yk - a ç å xk÷ èk=1 ø 35 - (-0.87)(0) b = k=1 =7 = 5 n

y = - 0.87 x+ 7 and the value ofy Therefore, the equation of the least squares line is will be estimated to y=-0.87(9)+7=-0.83.

7–27 ..

Calculus

Chapter 7

5. The U.S. Centers for Disease Control and Prevention (CDC) use a measure called body mass index (BMI) to determine whether a person is obese. A BMI between 25.0 and 29.9 is considered overweight, and a BMI of 30.0 or more is considered obese. The following table shows the percent of United States males 20 years and older who were obese in the years indicated, judging on the basis of BMI. (Source: CDC, May 19, 2010)

Year 2000 2002 2004 2006 2008

Percent Obese 27.5 27.8 31.1 33.3 32.3

Using 0 for 2000, find the least squares line and use it to predict the percent of males 20 years and older that will be obese in the year 2014.

Totals

x

y

xy

0 2 4 6 8 20

27.5 27.8 31.1 33.3 32.3 152

0 55.6 124.4 199.8 258.4 638.2

x2 0 4 16 36 64 120

The normal equations would be 120a+20b=638.2 20a + 5b = 152 The solutions for a and b are

æn ö æn öæn ö n ç å xkyk ÷- çå xkøçå yk÷ è k=1 ÷èk=1 ø 5(638.2) - (20)(152) = 0.755 a = k=1 = 2 (20)2 5(120) æn ö æn ö n ç å x2k÷-çå xk è k=1 ø è k=1 ÷ ø n æn ö å yk - a ç å xk÷ èk=1 ø 152 - (0.755)(20) b = k=1 =27.38 = 5 n

Therefore, the equation of the least squares line is y=0.755x+27.38and the value of y when x=14will be estimated to y=0.755(14)+27.38=37.95.Therefore, approximately 38 percent of the male population will be considered obese in the year 2014.

7–28 ..

Calculus

Chapter 7

6. The table gives the winning distance in the discus throw in the Olympic Games from 1980 to 2008.

Year 1980 1984 1988 1992 1996 2000 2004 2008

Distance (ft) 66.64 66.60 68.82 65.12 69.40 69.30 69.89 68.82

x = 0 for Use a graphing calculator to find the least squares line for the data, letting 1980. Estimate the winning distance in the discus throw in the Olympic Games of 2020.

Rewrite the table using x for the year and y for the distance.

x 0 4 8 12 16 20 24 28

Y 66.64 66.60 68.82 65.12 69.40 69.30 69.89 68.82

On your graphing calculator, go into the “STAT” mode and enter the x values to list 1 and the y values to list 2. Next go to the “CALC” mode within “STAT”. Option number 4 is “LinReg(ax+b). Hit the enter key twice for the following result:

7–29 ..

Calculus

Chapter 7

y = 0.111x + 66.514and the value Therefore, the equation of the least squares line is of y when x=40will be estimated to y = 0.111(40)x + 66.514 = 70.95Therefore, a distance of approximately 70.95 feet will be the winning distance in the year 2020.

7–30 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-6 Double Integrals over Rectangular Regions Goal: To evaluate indefinite and definite double integrals

Definition:

Double Integral

The double integral of a functionf (x,y) over a rectangle R = {( x,y ) | a £ x £ b,c £ y £ d} is béd f ( x,y)dy ù dx òòf(x,y)dA = òaëòc û R

ù dé b = òcëòa f ( x,y)dx dy û Definition:

Average Value over Rectangular Regions

The average value of the function f(x,y)over the rectangle R = {( x,y)|a£x£b,c£y£d}is 1 òòf(x,y)dA (b - a )(d - c ) R

In Problems 1 and 2, find each antiderivative. Then use the antiderivative to evaluate the definite integral.

1. a.

ò25x4y3dx

a. ò25x4y3 dx= 5x5 y3 +E(y)

b.

2 ò025 x4y3 dx

b.

2 5 3 ò025 x4y3 dx= 5 x y 0

2

= 5(2)5 y3 - 5(0)5 y3 =125 y3

7–31 ..

Calculus

Chapter 7

ln x ò 2 dy xy

b.

lnx lnx a. òxy2 dy = òx y-2 dy

b.

2. a.

2ln x ò1 2 dy xy 2

2 lnx

ò1xy2 dy = -

xy-1 ln = +C(x) x -1 ln x =+C(x) xy

lnx xy 1

lnx æ lnx ö - ç - ÷ x(2) è x(1 ø lnx ln x ) lnx =+ = 2x x 2x =-

In Problems 3 and 4, evaluate each iterated integral.

64 3. ò1ò-2(8x+6y+7)dxdy x =4 ù 6é 6 4 ò1 ò- 2 (8x + 6 y + 7) dxdy = ò1 ê4x2 +6xy+7xx =- ú dy 2û ë +6(4)y+ 7(4)) - (4(-2)2 + 6(-2) y + 7(-2)) ù dy é =16òë(4(4)2 û

- (16- 12y- 14)] dy = ò16 [(64 + 24 y + 28) 6 = ò1 [ 90+36y]dy 6

= 90y+18y2

1

= (90(6)+18(6)2)- (90(1)+18(1)2) 6 4

ò1 ò- 2 (8x + 6 y + 7) dxdy = - 216

7–32 ..

Calculus

Chapter 7

4 20x3 y4 dy dx 4. ò-3 2 ò0 y=4 ù 3 4 3 é 3 5 ò-2ò020x3y4dydx = -ò2 ê4x y y=0 ú dx ë û 3 é(4x3(4)5 ) (4 3(0)5) ù - x =ò û dx -2ë

= ò3 é(4(1024)x3) - (0) ùdx -2ë û 3 é ù dx 4096 x3û = ò-2 ë =1024 x4

3 -2 4

= (1024(3) ) - (1024(-2)4) 3 4 20 x3 y4 dydx = 66, 560 ò- 2 ò0

5. Use both orders of iteration to evaluate the double integral. òò(x3y3)dA;

R={(x,y)|-3£x£0,0£y£4}

R

1 4 é 4 3 x=0 ù 3 3 4 0 (x y ) dx dy = 4 òòò0 êë( x y ) x=- ú dy 0 3 3û 1 é 3y ) - ((-3)4 y3)ù = 4 04òë((0)4 û dy 4 81 y3 ù dy =òé0ë û 81 4 44

= - 16 y 0 81 (0)4) = (- 81 ( 4 )4 ) - (- 16 16

= - 1296 1 0 é 3 4 y=4 0 4 3 3 (x y ) dydx = 4 ò-3 ò- 3 ò0 êë( x y ) y=ù dx 1 é 0 0 (x 3(4) 4) - (x3(0) 4) ù = 4 ò-3 û dx ë ú 0 é x3ù 64 dx û = ò-3 û ë =16x4

0 -3

= (16(0)4 ) - (16(- 3)4 ) = - 1296 7–33 ..

Calculus

Chapter 7

6. Find the average value of the function over the given rectangle.

f(x,y)=3x2-4y3; R={(x,y)|0£x£4,-3£y£1} Using the average value formula yields the result x=4 ù 1 1 1 é 3 - 4xy3 x=0 1 4 (3x2 - 4 y3 ) dx dy= òêx ò- 3 ò0 ú dy 16 3ë (4- 0)(1- (- 3)) û

=

1 1 é64-16y3ù dy û 16 ò 3ë

=

1 (64y- 4y4)1

-3 16 1 = ((64- 4)- (-192- 324)) 16 36 =

7. Find the volume of the solid under the graph of the function over the given rectangle.

f(x,y)=4x+8y-7; R={(x,y)|-2£x£2,-1£y£3}

x=2 3 é 2 x=-2 ù 3 2 ò-1 ò-2 (4x +8 y - 7) dxdy = ò- 1ê 2 x +8 x y - 7 x ú dy ë û 3 dy] = ò-1[(8 +16y- 14)- (8 - 16 y + 14) 3 = ò-1[32y - 28]dy

=16y2 - 28y

3 -1

= (16(3)2 - 28(3))- (16(- 1)2 - 28(- 1)) =16cubicunits

7–34 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-7 Double Integrals over More General Regions Goal: To evaluate double integrals over non–rectangular regions

Definition:

Regular Regions

A region R in the xy plane is a regular x region if there exist functions f(x) and g(x) and numbers a and b such that

R={(x,y)|g(x)£y£f(x),a£x£b} A region R is a regular y region if there exist h(y) and k(y) and numbers c and d such that

R={(x,y)|h(y)£x£k(y),c£y£d} Definition:

Double Integration over Regular Regions

Regular x Region If R={(x,y)|g(x)£y£f(x),a£x£b},then f(x) F(x,y)dyù òòF(x,y)dA=òéb ëòg(x) û dx a R

Regular y Region IfR={(x,y)|h(y)£x£k(y),c£y£d},then

k(y) ù dy òòF(x,y)dA = òcd éëòh(y)F(x,y)dx û R

7–35 ..

Calculus

Chapter 7

1. Evaluate the following integral.

6 x ò1ò0(3x+y2)dydx

y3 6 ò1ò0(3x + y ) dy dx= ò1 (3xy + 3 ) 6 x

2

y=x

dx y=0

( x)3

(0)3

= ò16 (3x(x)+ 3 ) - (3 x (0)+ 3 )dx = ò16 (3x2 + 3x3)dx 4

6

x) = ( x3 + 12 ((6)3 1

(6) 4

(1)4

+ 12 ) - ((1)3 + 12 )

=

1) +1296 = (2592 ) - ( 1 + 12 12 12 6 x

2 ò1ò0(3x + y ) dy dx= 3886 » 323.83

12

12

2. Use the description of the regionR to evaluate the integral.

òò(2x+y-3)dA;

R={(x,y ) | 0 £x£y,0£y£6}

R 6 y 1 6 2 ò0ò0(2x + y - 3) dxdy = 4ò0( x + y-3) 1 6

x=y x=0

dy

2 y + y-3) 2 - (2(0) + y - 3)2 dy 4ò0 = 1 6 ( y - 3)2 - ( y - 3)2 dy 4ò0 = 1 6 y2 - 12)y dy 2 4ò0 6 8y3 = 14 ( 3(3 - 6 y2 ) =

(8 3

0

3

1 8(6)

1 8(0) = ( 3 - 6(6)2)- 4 ( 3 - 6(0) 2 ) 4 6 y ò0ò0(2x + y - 3) dxdy = 90

7–36 ..

Calculus

Chapter 7

3. Graph the region R bounded by the graphs of the indicated equations. Describe R in set notation with double inequalities, and evaluate the indicated integral.

y=1+6x-x2, y =1; òò15 xdA R

The region bounded by the equations is (the window shown is quadrant 1 only)

Therefore, the regionR is defined by R ={( x,y ) | 1 £ y £ 1 + 6x - x2 , 0 £ x £ 6}.

61+6 x-x 2

ò0 ò1

1 6x- x2 y=+

(15x)dydx = ò06 (15xy) y=1

dx

= ò06 (15x(1+6x - x2))- (15x(1))dx = ò06 (90x2 - 15x3) dx = (30x3 -

15x4 60

= (30(6)3 61+6 x-x 2

ò0 ò1

4

)

15(6)4 (30(0)3 - 15(0)4 ) 4 ) 4

(15x)dydx =1620

7–37 ..

Calculus

Chapter 7

4. Graph the region of integration, reverse the order of integration, and then evaluate the integral with the order reversed.

4 x ò0ò0(y-x)3dydx

The region of integration will look as follows:

Therefore, reversing the order of integration will yield x=4 1 4 ò0( y - x)4 y dy x= 4 1 4 = - ò0 ( y - 4) - ( y - y )4 dy 4 1 4 = - ò0 ( y - 4 dy 4 1 4) 4 = - 20 ( y - 4)5 0

4 4 ò0 òyy(- x)3 dxdy = -

4 4 òyy(- x)3 dxdy = ò0

1 ((0)5 4 5 256 - (- 4)) = 20 5

7–38 ..

Calculus

Chapter 7

5. Reverse the order of integration and then evaluate the integral with the order reversed.

Do not attempt to evaluate the integral in the original form.

25 4 ò0 ò y ( x3+4)dxdy

5 x2

3

ò0ò0 ( x

+ 4) dydx = ò05 y = ò05 2x x 52 = ò0

y=x2

x3 + 4

dx y=0

x 3 + 4 0- x3 + 4 dx x3 + 4 dx

3 5 1 = (x3 + 4) 20 3 3 3 1 = 1 ((5)3 + 4) 2 -3 ((0)3 + 4) 2 3 5 x2 3 ò0ò0 ( x + 4) dydx » 488.39

7–39 ..

Calculus

Chapter 7

6. Use a graphing calculator to graph the region R bounded by the graphs of the indicated equations. Use approximation techniques to find intersection points correct to two decimal places. Describe R in set notation with double inequalities, and evaluate the indicated integral correct to two decimal places.

y=x+3,y=x2,x=0;

òò12xydA R

The region bounded by the above equations is

The point of intersection is found by using the intersect feature on the calculator. The point of intersection is (2.30, 5.30). Therefore, the region of integration R is defined by R={(x,y)|x2£y£x+3,0£x£2.30}. y=x+3 2.30 x+3 (12xy)dydx = ò02.30 (6 xy2 ) y=x2 dx ò2 x

ò0

= ò02.30 (6 x( x + 3)2 )- (6x(x2 ) 2 ) dx +6x3 = ò02.30 ( - 6 x5 + 36x2 + 54 x) dx 2.30

x2 0 = - x6 +23 x4 +12x3 +27

= (-(2.30)6 + 23(2.30)4 +12(2.30)3 +27(2.30)2)(- (0)6 + 3(0)4 +12(0)3 +27(0)2) 2

2.30 x+3 ò0 òx2 (12xy)dy dx » 182.77

7–40 ..

Calculus

Chapter 8

Name ________________________________ Date ______________ Class ____________

Section 8-1 Trigonometric Functions Review Goal: To review the basic concepts involved with trigonometric functions

Conversion Factors:

1 =

 rad

1rad =

180

180



Sine and Cosine Functions with Real-Number Domains

Definition:

sinx =sin(xradians) and Definition:

cosx=cos(xradians)

Four Other Trigonometric Functions

tan x =

sin x , cosx¹0 cos x

secx=

1 , cosx¹0 cos x

cot x =

cos x x , sin x ¹ 0 sin

csc x =

1 , sin x ¹ 0 sinx

In Problems 1 and 2, convert the given degree measure to radian measure.

1.

210o

210o p 210p 7p 210 = 1 = = rad 180 180 6 o

2.

-225o o

-225 =

-225 o p -225p 5p = =rad 1 180 180 4 8–1 ..

Calculus

Chapter 8

In Problems 3 and 4, convert the given radian measure to degree measure.

3.

p 10

p 180o 180op p =18o = = 10 10 p 10p

4.

-

3p 4

-

3p -3p 180o -540op =-135o = = 4 4 p 4p

In Problems 5–10, find the exact value of each expression.

5.

sin30

1 Using Figure 8 on page 497 givessin 30 = 2 .

6.

cos45 Using Figure 8 on page 497 gives cos45=

7.

2 .

2

cos Using Figure 6 on page 496 givescos = - 1.

8.

æ2ö

sinç- ÷3ø è

æ 2ö 3 Using Figures 5 and 8 on pages 496 and 497, respectively gives sinç- ÷=. è 3ø 2

8–2 ..

Calculus 9.

Chapter 8

cos(-60) 1 Using Figures 5 and 8 on pages 496 and 497, respectively gives cos(-60)= .

2

10.

sin135 2 .

2 In Problems 11–14, use a calculator in radian or degree mode, as appropriate, to find the value of each expression to four decimal places.

Using Figures 5 and 8 on pages 496 and 497, respectively gives sin135=

11.

tan(-73o)

The angle measurement is in degrees, we must make sure that the calculator is in degree mode. Therefore, tan(-73o)=-3.2709.

12.

sin(2.13)

The angle measurement is in radians, we must make sure that the calculator is in radian mode. Therefore, sin(2.13)=0.8477.

13.

æ7pö secç ÷ è 8ø

The angle measurement is in radians, we must make sure that the calculator is in æ7p ö 1 1 radian mode. Therefore, secç ÷= = =-1.0824. è 8 ø cos(7p ) -0.923880 8

14.

cos(169o)

The angle measurement is in degrees, we must make sure that the calculator is in degree mode. Therefore, cos(169o)=-0.9816.

8–3 ..

Calculus

Chapter 8

8–4 ..

Calculus

Chapter 8

Name ________________________________ Date ______________ Class ____________

Section 8-2 Derivatives of Trigonometric Functions Goal: To find the first derivatives of the trigonometric functions

Derivatives: Basic Form

d sin x=cosx dx

d cos x = - sin x dx

d sec2 x dx tanx = d dx sec x = sec x tan x

d csc2 x dx cot x = d dx csc x = - csc xcot x

Derivatives: Generalized Form

For u=u(x),

d du sinu = cos u dx dx

d du cos u = - sin u dx dx

d sec2 u du tanu = dx dx d dx secu = secu tanu du dx

d csc2 u du cot u = dx dx d dx cscu = - csc ucotu du dx

8–5 ..

Calculus

Chapter 8

In Problems 1–6, find the indicated derivative.

1.

d (cos(5x+3)) dx The angle measurement is given in function form, therefore use the generalized form.

d du cos u = - sin u dx dx d d cos(5x +3) = sin(5x + 3) (5x + 3) dx dx + - sin(5x 3)(5)

2.

= = d (12tanx) dx

5 sin(5x + 3)

The angle measurement is given in basic form, therefore use the basic form.

d tan x = sec 2 x dx d 12 tanx =12sec2 x dx

3.

d dx

[7x(sin(7x+2))]

The function is a product and the trigonometric function has an angle measurement in function form. Therefore use the product rule and the generalized trigonometric derivative. d dx

[7x(sin(7x+2))]=7x(cos(7x+2)(7)+sin(7x+2)(7) =49xcos(7x+2)+7sin(7x+2)

8–6 ..

Calculus

4.

Chapter 8

d ætantö ÷ dt çèsectø

The function is a quotient and the trigonometric function has an angle measurement in basic form. Therefore use the quotient rule and the basic trigonometric derivatives.

d ætantö sect(sec2t ) - tant(sec ttant) ÷= 2 dt èç sect ø (sec t)

5.

=

sec3t - tan2tsect sec2t

=

sec2t - tan2t sect

d (tan(3pu) du The angle measurement is given in function form, therefore use the generalized form.

d (tan(3pu) = sec 2(3pu)(3p) du = 3psec2(3pu)

6.

d (sect)6 dt

The function is a power and the trigonometric function has an angle measurement in basic form. Therefore use the power rule and the basic trigonometric derivatives.

d (sect)6 6(sect)5(sec tan ) t t = dt = 6 tan t sec6 t

8–7 ..

Calculus

Chapter 8

8–8 ..

Calculus

Chapter 8

Name ________________________________ Date ______________ Class ____________

Section 8-3 Integration of Trigonometric Functions Goal: To evaluate indefinite and definite integrals that involve trigonometric functions

Integrals :

Indefinite Integrals of Sine and Cosine

òsinudu = - cos u + C

òcosudu = sin u + C

In Problems 1–5, find each indefinite integral.

1.

òcos(5x)dx

Let u=5x,therefore du = 5dx.Use the method of substitution and the integral of the cosine to solve the problem.

1 òcos(5x)dx = 5 òcos udu 1 = 5 (sin u + C ) sin(5 ) x = +C 5

8–9 ..

Calculus 2.

Chapter 8

ò(cosx)5sinxdx

Let u=cosx,therefore du=-sinxdx.Use the method of substitution to solve the problem.

ò(cosx)5sinxdx = - òu5du

==-

3.

u6

+C 6 (cos x)6 6

+C

ò(6x+5)sin(3x2+5x-7)dx

Let u=3x2+5x-7,therefore du= (6x + 5)dx.Use the method of substitution and the integral of the sine to solve the problem. ò(6x+5)sin(3x2 +5x- 7) dx = sin udu ò cos u + C = cos(3x2 + 5x - 7) + C = 4.

sin5x dx ò cos5x

-

Let u = cos(5 x ),therefore du = - 5sin(5 x) dx.Use the method of substitution to solve the problem.

1 1 sin5x dx = - 5 òu du ò cos5x 1 = - 5 ln u + C 1 5 = - ln cos5x + C

8–10 ..

Calculus

5.

Chapter 8

cscx dx ò secx Rewrite the problem in terms of the sine and cosine functions.

cscx cosx dx dx=ò ò secx sinx Let u=sinx,therefore du=cosxdx.Use the method of substitution to solve the problem.

cos x 1 dx = ò du ò sin x u = ln u + C = ln sin x + C

In Problems 6–8, evaluate each of the definite integrals. 

6.



ò02cosxd x = sin x2 0 = sin - sin0 2

=1 - 0 

2 xdx =1 ò0cos

7.

3 2 sin xdx =

ò

3

cos x 2 4

4

-

cos(3) - (- cos(  )) 2

4

= 0 - (-0.7071) = 3 = ò2 sin xdx 0.7071 4

8–11 ..

Calculus

8.

Chapter 8

7

7 ò3sinxdx = cos x 3

- cos7-( - cos 3) = 0.753902 - (- 0.989992) = 7 ò3sin xdx -» 1.7439

Be sure that your calculator is in radian mode to solve this problem.

8–12 ..