M x ( R 2 ) = ∫ a (( ( )) gx − ( ( )) ) fx dx

1 7 M x ( R 2 ) = ∫ a (( ( )) gx − ( ( )) ) fx dx

M x = M x ( R 1 ) + M x ( R 2 ) = δδ += δ

M y ( R 1 ) = δ ∫ a xhx (() − ( )) g x dx

Therefore, the centroid is given by

= δ ∫ (() a hx − gx () + gx () − ( )) f x dx

a (() hx − ( )) g x dx + δ ∫ a (() gx − ( )) f x dx

= mR ( 1 ) + mR ( 2 )

19. mR ( 1 ) = δ ∫ (() gx − ( )) f x dx

(( ( )) 2 − ( ( )) ) c 2 M x R 3 ∫

∫ 2 − ( )) f x dx

(( ( )) hx δ 2 b 2 − ( ( )) gx 2 + ( ( )) gx 2 − ( ( )) ) fx 2 dx

M x ( R 1 ) = ∫ (( ( )) gx − ( ( )) ) fx 2 dx

M y ( R 2 ) = δ ∫ xgx (() − ( )) f x dx

= δ xhx (() − gx () + gx () − ( )) ∫ f x dx (() gx − ( ))

a f x dx

= δ ∫ a (() gx − ( )) f x dx + δ ∫ b (() gx − ( )) f x dx

= δ ∫ a xhx (() − ( )) g x dx + δ ∫ a xgx (() − ( )) f x dx

= mR ( 1 ) + mR ( 2 )

M x ( R 3 ) = ∫ (( ( )) gx − ( ( )) )

fx dx

21. Let region 1 be the region bounded by x = –2, x = 2, y = 0, and y = 1, so m 1 =⋅=. 41 4

1 By symmetry, x 1 = and 0 y 1 = . Therefore

(( ( )) gx ∫ 2 − ( ( )) ) fx 2 dx

M 1 y 2 = xm 11 = 0 and M 1 x = ym 11 =. 2

Let region 2 be the region bounded by x = –2, x = 1, y = –1, and y = 0, so m 2 =⋅=. 313

M y ( R 3 ) = δ ∫ xgx (() − ( )) 1 a 1 f x dx By symmetry, x 2 = − and y 2 = − . Therefore

Instructor’s Resource Manual

Section 5.6 331 Section 5.6 331

region 3 be the remaining region, so m 3 = 22 .

3 symmetry, x = − and 1 y 1 = . Therefore,

2 By symmetry, x 3 = 2 and y 3 = − . Therefore,

M 1 y = xm 1 1 =− 20 and M 1 x = ym 1 1 = 30 . Let

M 3 y = xm 3 3 = 44 and M 3 x = ym 3 3 =−. 11 region 2 be the region bounded by x = –3, x = –2,

= . By symmetry, M 1 y + M

x 2 = − and y 2 = − . Therefore, 2

region 3 be the region bounded by x = 0, x = 1,

1 y = –2, and y = –1, so m 3 = . By symmetry, 1 1 ⎡ 1 ⎤

25. A = x dx 3 =

x 3 = and y 3 = − . Therefore,

From Problem 11, x =.

m 1 + m 2 + m 3 23 46 Using cylindrical shells:

x y M 2 3 x 2 49 V ∫ 0 x x dx

x dx

m 1 m 2 + m 3 23 46

26. The area of the region is π . The centroid is the a 2 x = 2, y = 2, and y = 4, so m 1 = ⋅ = . By 42 8 center (0, 0) of the circle. It travels a distance of

23. Let region 1 be the region bounded by x = –2,

symmetry, x 1 = 0 and y 1 = . Therefore, 3 2 π (2a) = 4π a. V =π 4 23 a M 1 y = xm 11 = 0 and M 1 x = ym 11 = 24 . Let

27. The volume of a sphere of radius a is π 3 a . If

region 2 be the region bounded by x = –1,

x = 2, y = 0, and y = 2, so m 2 = ⋅ = . By 32 6

the semicircle

y = a − x is revolved about symmetry, x 2 = and y 2 = . Therefore, 1 the x-axis the result is a sphere of radius a. The

centroid of the region travels a distance of 2 π y .

M 2 y = xm 2 2 = 3 and M 2 x = ym 2 2 = . Let 6

region 3 be the region bounded by x = 2, x = 4,

The area of the region is π a . Pappus's

y = 0, and y = 1, so m

3 = ⋅ = . By symmetry, 21 2 Theorem says that

x 3 = 3 and y 2 = . Therefore, M 3 y = xm 33 = 6 ⎛ 1 2 ⎞

22 (2 4 π y ) π a =π ay =π a 3

and M 3 x = ym 33 =. 1

, x = (by symmetry) 0

m 1 + m 2 + m 3 16

2 x + M 3 x 31 = 28. Consider a slice at x rotated about the y-axis. m b

1 + m 2 + m 3 16 Δ= V 2 π xh x () Δ , so x V =π 2 ∫ xh x dx () . a

24. Let region 1 be the region bounded by x = –3, b Δ≈ m hx () Δ , so x m = ∫ () h x dx = A .

symmetry, x 1 = − and 2 y 1 = − . Therefore,

M 1 y = xm 1 1 =− 12 and M 1 x = ym 1 1 = − . Let 3 M y ∫ xh x dx ()

region 2 be the region bounded by x = –1, x = 0,

y = –2, and y = 0, so m 2 = . By symmetry, 2 The distance traveled by the centroid is 2 x π.

2 = − and y 2 = − . Therefore, 1 ∫ a

(2 π xA ) =π 2 xh x dx ()

Therefore, V =π 2 xA .

Section 5.6 Instructor’s Resource Manual

π r cos , so the centroid travels a distance

V =π 2 ∫ c ( K − y w y dy )()

of 2 π r cos

b. Δ≈ m wy () Δ , so y m = ∫ c w y dy () = A .

Thus, by Pappus's Theorem, the volume of the resulting solid is

M x ≈ yw y () Δ , so y M x = ∫ yw y dy () .

∫ yw y dy ()

y = c =π 4 rn sin cos

The distance traveled by the centroid is

c Kw y dy ∫ c yw y dy () ⎞ ⎟

=π d 2 ∫ ( K − y w y dy )() As n → ∞ , the regular polygon approaches

a circle. Using Pappus's Theorem on the

Therefore, V =π 2( K − yA ) .

circle of area π whose centroid (= center) r 2

travels a distance of 2 πr, the volume of the

30. a. m = bh solid is ( 2

π r )(2 π=π r ) 2 23 r which agrees

b with the results from the polygon.

The length of a segment at y is b − y .

32. a. The graph of (sin ) f x on [0, π ] is

Δ M x ≈ yb ⎜ − y ⎟ Δ= y ⎜ by − y ⎟ Δ y

π symmetric about the line x = since

x = ∫ by − y dy 0 ⎜ ⎟ ⎝

f (sin ) x = f ( sin( π− x ) ) . Thus x =.

⎢ by −

y 3 ⎤ = bh 2 π

⎣ 2 3 h ⎦ ⎥ 0 6 ∫ xf (sin ) x dx x π = 0

0 (sin ) f x dx

Therefore

b. A = bh ; the distance traveled by the

∫ xf (sin ) x dx = 0 2 ∫ (sin ) f x dx

centroid is 2 k − ⎞ ⎜ .

31. a. The area of a regular polygon P of 2n sides

. (To find this consider

the isosceles triangles with one vertex at the center of the polygon and the other vertices on adjacent corners of the polygon. Each

such triangle has base of length 2 sin r

2 n and height cos r π⎞ .

2 n ⎟ Since P is a regular ⎠ polygon the centroid is at its center. The distance from the centroid to any side is

Instructor’s Resource Manual

Section 5.6 333

1 1 ⎡ centered inside the hole. We then recompute the

g ∫ 2 () x − f 2 () x ⎤ dx

y centroid of Problem 34 (in this position) as SR − y R

∑ xh ii

20 f x dx

R ∫ g () x − 2 f () x ⎤ dx ≥ ( S − R ) 2 ∫ ()

0 f x dx i 2 = 2 ( 25)(6.5) − +− ( 15)(8) + + (5)(10) + (10)(8)

R g () x − f ()

2 2 ∫ 0 x dx

1 1 2 1 1 2 2 ∑ (( h i − 4) −− ( 4) )

R ∫ g 0 () x dx ≥ S ∫ () 0 f x dx

2 ∫ g () 1 0 2 x dx 2 ()

34. To approximate the centroid, we can lay the

figure on the x-axis (flat side down) and put the

shortest side against the y-axis. Next we can use the eight regions between measurements to

A quick computation will show that these values

agree with those in Problem 34 (using a different height of the ith region, be approximated by the

approximate the centroid. We will let h i ,the

reference point).

height at the right end of the interval. Each Now consider the whole lamina as R 3 , the interval is of width Δ = cm. The centroid can x 5 circular hole as R 2 , and the remaining lamina as

be approximated as R 1 . We can find the centroid of R 1 by noting

and similarly for M y ( R 1 ) ∑ .

From symmetry, we know that the centroid of a circle is at the center. Therefore, both

23.38 M x ( R 2 ) and M y ( R 2 ) must be zero in our case.

This leads to the following equations

() h i 2

2 (1/ 2)(6.5 2 + 8 2 ++ 10 2 + 8) 2 y 3 y i 2 1 x =

mR ( 3 ) − mR ( 2 )

Thus, the centroid is 7 cm above the center of the hole and 0.669 cm to the right of the center of the hole.

Section 5.6 Instructor’s Resource Manual Section 5.6 Instructor’s Resource Manual

+ 0.05 0.05 0.05 + = 0.15 this problem, it will be necessary, in some

regions, to find the value of g(x) instead of just

f (x) – g(x). We will use the 19 regions in the

b. EX () = ∑ xp ii

figure to approximate the centroid. Again we

choose the height of a region to be approximately = 0(0.7) 1(0.15) 2(0.05) + + the value at the right end of that region. Each

+ 3(0.05) 4(0.5) + region has a width of 20 miles. We will place the

north-east corner of the state at the origin.

The centroid is approximately

3. a. PX ( ≥ 2) = P (2) = 0.2

∑ xfx i (() i − gx ( )) i

b. EX () =− 2(0.2) ( 1)(0.2) 0(0.2) +− +

b. EX () =− 2(0.1) ( 1)(0.2) 0(0.4) +− +

∑ (() fx i − gx ( )) i i

b. EX ( ) 1(0.4) 2(0.2) 3(0.2) 4(0.2) = + + +

6. a. PX ( ≥ 2) = P (100) + P (1000) This would put the geographic center of Illinois

= 0.018 0.002 + = 0.02 just south-east of Lincoln, IL.

b. EX () =− 0.1(0.98) 100(0.018) +

5.7 Concepts Review

1. discrete, continuous

7. a. PX ( ≥ 2) = P (2) + P (3) + P (4)

2. sum, integral 3 2 1 = 6 + + = = 0.6 10 10 10 10

3. ∫ () f x dx

0 b. EX ( ) 1(0.4) 2(0.3) 3(0.2) 4(0.1) = + + + = 2

4. cumulative distribution function

Problem Set 5.7

1. a. PX ( ≥ 2) = P (2) + P (3) =

b. EX () = ∑ xp ii

i = 1 = 0(0.8) 1(0.1) 2(0.05) 3(0.05) + + +

Instructor’s Resource Manual

Section 5.7 335

10 10 10 10 3 = 20 ∫ ( 20 x 2 − x 3 ) dx

3 ⎡ 20 x 3 x 4 ⎤

b. EX () = 0(0.4) 1(0.1) 2(0) 3(0.1) 4(0.4) + + + + 20

9. a. PX ( ≥ 2) = ∫ dx = ⋅ 18 = 0.9

2 20 20 c. For 0 ≤≤ x 20

20 Fx () = ∫ 0 t (20 − t dt )

20 1 ⎡ x 2 ⎤

b. EX () = ∫ x ⋅ dx = ⎢ ⎥ = 10 x

c. For between 0 and 20, x

0 13. a. PX ( ≥ 2) = ∫ x (4 − x dx ) 20 20 20 2 64

Fx 4 () = ∫ dt = ⋅= x 3 2

34 ⎡ x 3 20 4 1 1 x 4 ⎤

10. a. PX ( ≥ 2) =

b. EX () = ∫ x ⋅ dx = ⎢ ⎥ =−= 55 0 b. EX () =

∫ x ⋅ x (4 − x dx ) 0

∫ 4 x − x ) dx = ⎢ x 0 − ( ⎥ = 2.4 64 64

c. −≤≤

For 20 x 20 ,

Fx () = ∫

Fx () = ∫ t 2 (4 − t dt ) =

(8 − x dx ) ∫ x ⋅ x

14. a. PX ( ≥ 2) =

0 (8 − x dx ) 256 ∫ 2 32

0 b. EX () = ∫ x ⋅ (8 − x dx ) 0

32 3 ∫ 3 t (8 − t dt ) = ⎢ 4 t − ⎥ ⎢ ⎣ ⎥

3 2 1 c. For 0 ≤≤ x 8

Fx () = ∫ (8 − t dt ) = ⎢ 8 t − ⎥

12. a. 0 PX ( ≥ 2) = ∫ x (20 − x dx )

Section 5.7 Instructor’s Resource Manual

15. a. PX ( ≥ 2) = ∫ sin

dx

b. EX () = ∫ x ⋅ 3 dx =− ⎢ ⎥ = 1.8

∫ 1 40 3 t ⎢ ⎣ 80 t 2 ⎥

b. EX () = x ⋅ sin

dx ∫ ⎦

81 81 81 x 2 − 81

Using integration by parts or a CAS,

EX () =

80 80 x

19. Proof of ( ) Fx ′ = fx ():

c. For 0 ≤≤ x 4

By definition, ( ) Fx = A ( ) . By the First f t dt

Fx () =

sin

∫ 0 8 ⎜ 4 ⎟ dt =

cos

4 ⎦ ⎥ 0 Fundamental Theorem of Calculus,

2 4 2 Proof of ( ) FA = 0 and ( ) 1: FB =

FA () =

16. a. PX ( ≥ 2) = ∫ cos ⎜ ⎟ dx ∫ A

() f x dx = 0;

4 FB () = ∫ A () f x dx = 1

⎢ sin ⎜

⎟ ⎥ = sin − sin =− 1 Proof of ( Pa ≤ X ≤ b ) = Fb () − Fa ():

Pa ( ≤ X ≤ b ) = ∫ a () f x dx = Fb () − Fa ( ) due to

b. EX () =

∫ the Second Fundamental Theorem of Calculus. x ⋅ cos dx

Using a CAS, EX ( ) 1.4535 ≈ a + 20. a. b

The midpoint of the interval [a,b] is .

Fx () = ∫ cos ⎜ ⎟ dt = ⎢ sin ⎜ ⎟ ⎥

17. a. PX ( ≥ 2) = ∫ 2 dx =− ⎢ =

b. EX () = ∫ x ⋅

dx =

b. EX () = ∫ x ⋅

c. Fx () = ∫

dt =

c. For 1 ≤≤ x 4 a b − a b − a b − a

Fx () ∫ dt ⎢ ⎥ = + 21. 1 The median will be the solution to the

a = x 0.5 − ∫ .

equation 0 dx

x 0 − a ) = 0.5

18. a. PX ( ≥ 2) = ∫ 2 3 dx =− ⎢ ⎥ x 0 −= a

Instructor’s Resource Manual

Section 5.7 337

22. The graph of fx () =

x 2 (4 − x ) 2 is

c. EX () = ∫ x ⋅ (2 −− x 2) dx

symmetric about the line x = 2. Consequently,

x ⋅ (2 ( +− x 2)) dx + ∫ x ⋅ (2 ( −− x 2)) dx

PX ( ≤ 2) = 0.5 and 2 must be the median of X.

= ∫ x dx + ∫ (4 x − x 2 dx

23 . Since the PDF must integrate to one, solve

0 d. x If 0 ≤≤ x Fx 2, ( ) = ∫ t dt = ⎢⎥ =

If 2 <≤ x Fx 4, ( ) = ∫ x dx + (4 − t dt )

x 2 125 ⎞ 3

+ ⎢ 4 t − ⎥ =+ ⎜ ⎜ x −

24. Solve kx (5 2 ∫ − x ) 2 dx = 1 x 2

+− x 1

k ∫ ( 25 x − 10 x + x

e. Using a similar procedure as shown in part

(a), the PDF for Y is

( 120 −− y 120 ) 0

25. a. Solve 2 ∫ k ( −− x 2 ) dx = 1 fy () =

Due to the symmetry about the line x = 2, the

solution can be found by solving

If 0 ≤< y 120, ( ) Fy = ∫

t dt

2 ∫ kx dx = 1

kx ⋅ = 1

4 k = 1 If 120 <≤ y 240,

Fy () =+

(240 − t dt

b. P (3 ≤ X ≤ 4) = ∫ ( 2 −− x 2 ) dx =+

Section 5.7 Instructor’s Resource Manual

26. a. Solve (180 2 ∫ kx − x dx ) = 1.

⋅ 6 ⎛ 180 3 Solve ∫

− ⎞ y ⎟ dy = 1 using a

⋅ ∫ 6 kt ⎜ − t ⎟ dt

b. P (100 ≤ X ≤ 150) 0 ⎝ 127 ⎠

1 2 Using a CAS,

x (180 − x dx )

F Y () y ≈ (7.54731 10 ) × 27 y 7 ( y 8 − 0.202475 y 7

+ 0.01802 y − 0.000923 y

y + (8.108 10 ) × − 9 y 2 − (6.156 10 × − 11 ) y

c. EX () =

∫ 2 x ⋅ x (180 − x dx )

0 28. a. Solve (200 2 ∫ kx − x 8 0 ) dx = 1.

Using a CAS, k ≈ 2.417 10 × − 23

27. a. Solve (0.6 6 ∫ 8 kx − x ) dx = 1.

b. The probability that a batch is not accepted is

0 (0.6 x ) dx 1 PX 200 2 ( ≥ 100) = k ∫ x (200 − x ) 8

dx Using a CAS, k ≈ 95,802,719

≈ 0.0327 using a CAS.

b. The probability that a unit is scrapped is

1 200 − P (0.35 ≤ X ≤ 0.45)

c. EX () = k ∫ xx ⋅ 2 (200 − x ) 8 dx

=− 1 k

∫ 0.45 6 x (0.6 − x ) 8 dx

50 using a CAS

≈ 0.884 using a CAS

d. Fx () = ∫ (2.417 10 × − ) (200 t 23 2 − t ) 8 dx

c. EX () =

x kx ⋅ 6 (0.6 − x ) 8 dx Using a CAS, F(x) ≈ (2.19727 10 × − 24 ) x ∫ 3

x = 8 k 8 ( − 1760 x 7 + 136889 x 6 − (6.16 10 ) × 8 x ∫ 5 x (0.6 − x ) dx

+ (1.76 10 ) × 11 x 4 − (3.2853 10 ) × ≈ 13 0.2625 x 3 + (3.942 10 ) × 15 x 2 − (2.816 10 ) × 17 x

Using a CAS,

Fx () ≈ 6,386,850 x 7 ( x 8 − 5.14286 x 7 e. Solve ∫ kx (100 − x ) dx . Using a CAS, 0

+ 11.6308 x 6 − 15.12 x 5 + 12.3709 x 4 k =

− 6.53184 x 3 + 2.17728 x 2 Fx y () = ∫ × − 20 t 2 − t 8 dt

0.419904 x + 0.36)

Using a CAS,

e. If X = measurement in mm, and Y =

Fx () ≈ (4.5 10 × − 21 ) x 3

measurement in inches, then Y = X / 25.4 .

x Thus, 8 ( − 880 x 7 + 342, 222 x 6 − (7.7 10 ) × 7 x 5

F Y () y = PY ( ≤ y ) = PX ( / 25.4 ≤ y )

+ (1.1 10 ) × 10 x 4 − (1.027 10 ) × 12 x 3

= PX ( ≤ 25.4 y ) = F ( 25.4 y )

+ (6.16 10 ) × 13 x 2 − (2.2 10 ) × 15 x

where Fx () is given in part (d).

Instructor’s Resource Manual

Section 5.7 339

From Problem 20, the CDF for X is Fx () = x ⎪

⎪ 0.95 if 2 ≤< x 3

Y is the distance from ( 1, X ) to the origin, so

⎪⎩ 1 if x > 3

Y = ( 10 − 2 +

Fx ()

X − 0 ) = 1 + X 2 1.0

Here we have a one-to-one transformation from

the set { x :0 ≤≤ x 1 } to { y :1 ≤≤ y 2 } . For

every 1 <<< a b 2 , the event a << Y b will

occur when, and only when, 0.4

a 2 −< 1 X < b 2 −

1 0.2 . If we let a = 1 and b = y , we can obtain the

0 1 2 3 x CDF for Y.

2 ( 2 1 ≤≤ Y y ) = P 1 −≤ 1 X ≤ y − 1 ⎧ 0 if x < 0

⎪ 0.7 if 0 ≤< x 1

⎪ ⎪ 0.85 if 1 ≤< x 2

34. Fx () =⎨

( ) 1 ⎪ 0.95 if 3 ≤< x 4

0.9 if 2 ≤< x 3

= F y 2 −= 1 y 2 −

To find the PDF, we differentiate the CDF with

⎪⎩ 1 if x ≥ 4

respect to y.

Therefore, for 0 ≤≤ y 2 the PDF and CDF are

respectively 0.6 y

gy () =

and Gy () = y 2 − 1 .

0. Consequently, 0 1 2 3 4 5 ∫ x x

() f t dt =

PX ( < c ) = PX ( ≤ c ). As a result, all four

35. a. PY ( < 2) = PY ( ≤ 2) = F (2) 1 =

expressions, ( Pa < X < bPa ), ( ≤ X ≤ b ),

Pa ( < X ≤ b ) and ( Pa ≤ X < b ), are

b. P (0.5 << Y 0.6) = F (0.6) − F (0.5)

31. By the defintion of a complement of a set, A ∪ A c = S , where denotes the sample space. S

Since ( PA ∪ A c ) = PA () + PA c ( ),

d. EY () = ∫ y ⋅

For Problem 1, 1 − PX ( <=− 1) 1 PX ( = 0)

=− 1 0.8 = 0.2 For Problem 2, 1 − PX ( <=− 1) 1 PX ( = 0) =− 1 0.7 = 0.3 For Problem 5, 1 − PX ( <=−= 1) 1 0 1

Section 5.7 Instructor’s Resource Manual

Concepts Test

b. P (1 << Z 2) = P (1 ≤≤ Z 2) = F (2) − F (1) 1. False: ∫ 0 cos x dx = 0 because half of the area

lies above the x-axis and half below the x-

axis.

2. True: The integral represents the area of the

c. fz () = Fz ′ () = ,0 ≤≤ z 3 region in the first quadrant if the center of

the circle is at the origin.

3. False: The statement would be true if either

d. EZ () = ∫ z ⋅ dz =

⎥ 2 f (x) ≥ g(x) or g(x) ≥ f(x) for

b. Consider Problem 1 with f(x)

= cos x and g(x) = 0.

37. EX () = ∫ x ⋅ x 2 (4 − x ) 2 dx = 2

4. True: The area of a cross section of a cylinder will be the same in any plane parallel to

2 4 and E(X ) 15 = ∫ x 2 ⋅ x 2 (4 − x ) 2 dx

the base.

5. True: Since the cross sections in all planes = ≈

4.57 using a CAS

parallel to the bases have the same area, the integrals used to compute the volumes

will be equal.

38. EX ( 2 ) = x 2 ∫ ⋅ x (8 − x dx ) = 19.2 and

6. False: The volume of a right circular cone of

3 8 x 3 EX 3 ( ) = ∫ ⋅ x (8 − x dx

1 radius r and height h is π rh 2 . If the

3 using a CAS

radius is doubled and the height halved

39. VX () = E ⎡ ( X − μ ) 2 ⎤ , where μ = EX =

the volume is π rh 2 .

x 2 (4 − x ) 2 dx = 7. ∫ False: Using the method of shells,

V =π 2 ∫ x ( − x 2 + x dx ) . To use the

40. μ = EX () = ∫ 0 x ⋅ x ( 8 − x dx = ) method of washers we need to solve 4

y =− x 2 + x for x in terms of y.

VX () = ∫ ( x − 4) ⋅ x (8 − x dx ) =

5 8. True: The bounded region is symmetric about

the line x =

41. E ⎡ X − μ 2 2

. Thus the solids obtained

⎢ 2 ⎣ ⎦ ⎥ = EX ( − 2 X μμ + )

by revolving about the lines

= EX ( 2 ) − E ( 2 X μ ) + E ( μ 2 )

x = 0 and x = 1 have the same volume.

= EX ( 2 )2 − μ ⋅ EX () + μ 2

9. False: Consider the curve given by x = ,

cos t

= EX ( 2 )2 − μ 2 + μ 2

since ( ) EX = μ

For Problem 37, ( ) VX = EX ( 2 ) − μ 2 and

using previous results, ( ) VX = 32 − 2 2 = 4 10. False: The work required to stretch a spring 2

7 7 inches beyond its natural length is

kx dx = 2, k while the work required to

stretch it 1 inch beyond its natural length

is ∫ kx dx = k 0 .

Instructor’s Resource Manual

Section 5.8 341

2. V =π ( x − x point downward, then the amount of 22 ∫ ) dx

x 2 − 2 x 3 the bottom of the tank is much less than 4 ∫ + x ) dx

water that needs to be pumped from near

the amount that needs to be pumped from

near the bottom of the cylindrical tank.

=π x 3 − x 4 + x 5 ⎤

12. False: The force depends on the depth, but the force is the same at all points on a surface

3. V =π 2 2 2 ∫ 3 xx ( − x ) dx =π 2 ∫ ( x − x dx )

as long as they are at the same depth.

13. True: This is the definition of the center of 1 ⎡ 1

=π 2 ⎢ x − x

mass.

14. True: The region is symmetric about the point

4. V =π ⎡ ( x − x 2 + 2) 2 − (2) ( π , 0). 2 ⎤ ∫ ⎣ ⎦ dx

15. True: By symmetry, the centroid is on the line

x 4 − 2 x 3 − 3 x ∫ 2 + 4) x dx

π x = , so the centroid travels a distance

16. True: At slice y, Δ≈ A (9 − y 2

x − 4 x 2 + 3) x dx

17. True: Since the density is proportional to the

square of the distance from the midpoint,

=π 2 ⎢ 4 x − x 3 + x 2 =

equal masses are on either side of the

midpoint.

1 2 1 3 1 4 18. True: See Problem 30 in Section 5.6. 1

xx ( − x ) dx ⎡ 3 x − 4 x ∫ ⎤ ⎣ ⎦ 1

1 True: A discrete random variable takes on a

6. x = 0 1 0 = =

∫ ( x − x 2 ) dx ⎡ 1 x 2 − 1 0 ⎤ ⎣ 2 x 3 finite number of possible values, or an 2 3 ⎦ 0

infinite set of possible outcomes provided

1 that these outcomes can be put in a list 1 22 11 ⎡ x 3 − 1 x 4 + 1 1 5 ⎤

20 ∫ ( x − x ) dx 23 ⎣ 2 5 x such as {x ⎦ 1 ,x 2 , …}. y =

2 ⎡ 1 ∫ 1 ( x − x ) dx

20. True: The computation of E(X) would be the

same as the computation for the center of

mass of the wire.

21. True: EX () =⋅= 515 1

7. From Problem 1, A = .

22. True: If ( ) Fx = ∫ f t dt , then Fx ′ () = fx

A ()

From Problem 6, x = and y = .

by the First Fundamental Theorem of

23. True: PX ( == 1) P (1 ≤ X ≤= 1) ∫ f x dx =

Sample Test Problems

2 2 ∫ 3 ( x − x ) dx = ⎢ x − x =

Section 5.8 Instructor’s Resource Manual

(4 x ∫ 2 − x ∫ 3 ) dx

x (4 x − x 2 ) dx

1 12. x =

a. W ∫ 2 x dx = x 2 = (64 4) − 4 2 ⎢ 32 x −

2 ∫ 0 (4 x ) dx

= 30 in.-lb

= 8 in.-lb 32

20 ∫ ⎣ (4 ) x − ( x 22 ) ⎤ 6 dx ⎦

9. W = (62.4)(5 ) (10 2 π − y dy ) y =

0 ∫ (4 x − x 2

0 ) dx

0 2 20 ∫ (16 x 2 − x 4 ) ⎣ dx ⎦ 0

= 1560 π ∫ (10 − y dy ) = 1560 π ⎢ 10 y − y 2 ⎥ 1 4

= 65,520 π 205,837 ft-lb 32

0 32 the object to the top without the cable and the

10. The total work is equal to the work W to pull up

3 W 5 to pull up the cable. 3 W 1 = 200 100 ⋅ = 20,000 ft-lb

work 2

120 6 13. V ∫ (4 ) x − ( x 22 ) ⎤ dx

The cable weighs

lb/ft.

(16 x 2 − x 4 ) dx

W 2 =Δ⋅= 6 yy 6 yy Δ

W 2 = ∫ 0 y dy

y =⎢ 2 ⎥

⎦ 0 Using Pappus’s Theorem:

= 6000 ft-lb

From Problem 11, A = 32 .

W = W 1 + W 2 = 26,000 ft-lb

From Problem 12, = 11. a. To find the intersection points, solve 32 y . 5

4x = x 2 .

=π⋅=π ⎛ ⎞⎛ ⎞ V 2048 2 yA 2

14. a. (See example 4, section 5.5). Think of

4 4 cutting the barrel vertically and opening the

0 (4 x x ) dx = ⎢ 2 x − x

lateral surface into a rectangle as shown in

3 ⎥ ⎦ 0 the sketch below.

b. To find the intersection points, solve

At depth 3 – y, a narrow rectangle has width y 2 − 16 y = 0 16 π , so the total force on the lateral surface

y (y – 16) = 0

is (δ = density of water = 62.4 lbs ft 3

A = ∫ ⎜ y − ⎟ dy = ⎢ y − y 2 ∫

δ (3 − y )(16 ) π dy = 16 πδ (3 − y dy )

− 32 = = 16 πδ ⎢ 3 y − ⎥ = 16 πδ (4.5) 14,114.55 lbs. ⎜ ≈

Instructor’s Resource Manual

Section 5.8 343

22. L 1 = ∫ a 1 + [ fx ′ () 2 ] dx

the same depth; thus the total force on the

bottom is simply the weight of the column of

L 2 = ∫ a 1 + [ gx ′ () ] dx

water in the barrel, namely

F = π (8 )(3) 2 δ ≈

37, 638.8 lbs. L 3 = fa () − ga () L 4 = fb () − gb ()

dy

15. = x 2 −

Total length = L 1 + L 2 + L 3 + L 4

dx

23. A 1 =π 2 ∫ a fx ()1 + [ fx () ] dx

dx

A =π 2 gx ()1 + gx ′ () dx

= ∫ x ++

dx = ∫

Total surface area = A 1 + A 2 + A 3 + A 4 .

24. a. PX ( ≥= 1) P (1 ≤ X ≤ 2)

= ∫ 8 − x ( 3 ) dx = 8 x −

The loop is − 3 ≤≤ t 3 . By symmetry, we can

b. P (0 ≤ X < 0.5) = P (0 ≤ X ≤ 0.5)

double the length of the loop from t = 0 to

dx

dy

∫ 0 12 ( 8 − x ) dx = ⎢ 8 x − 12 ⎥

L = 2 2 ∫ 3 t 4 + 2 t + 1 dt =

0 2 ∫ 0 ( t 2 + 1) dt = 85 ≈ 0.332

0 x c. EX () = x ⋅ 8 − x 3 ∫ 2

(9 − x 2 = 0.8 ∫ − 3 ⎜ ⎟ ∫ − 3 ) dx

a () ]

18. A = ∫ [ fx () − g x dx 12 ⎜ 4

20. V =π 2 ∫ ( x − a ) [ fx () − () g x dx ] b. fx () = Fx ′ () =⋅ 2 (6 − x ) =

y = δ ∫ xfx a [ () − () g x dx ] 6 ⎛ 6 − x ⎞

c. EX () = ∫ x ⋅⎜ ⎟ dx

∫ ⎡ f x − g 2 x ⎤ dx

Section 5.8 Instructor’s Resource Manual

Review and Preview Problems

10. a. ( 1 + 1 ) 1 = 2 1 = 2

1. By the Power Rule

=−+ C b. ( + 1 ) 10 ⎛ 11 1 ⎞ = ⎜ ⎟ ≈ . 2 593742 ∫ ∫

2. 100 By the Power Rule 1 101

3. By the Power Rule

dx = ∫ x dx =

1 10 10 By the Power Rule ⎛ 1 ⎞ ⎛ 11 ⎞

∫ . 0 99 dx = ∫ x 0 99 dx =

5. F () 1 = ∫ 1 dt = 0

6. By the First Fundamental Theorem of Calculus

Fx ′

1 dt = ≈ . dx 2 70481 ∫ t x

7. Let gx () = x 2 ; then by the Chain Rule and

DFx x ( ) = DFgx x ( ( )) = Fgxgx ′ ( ( )) ( ) ′

8. Let hx () = x 3 ; then by the Chain Rule and

b. ( 1 + ) 2 = () 12 . ≈ . 2 48832

problem 6,

D dt = DFhx ( ( )) = Fhxhx ′ ( ( )) ( ) ′

9. a. ( 11 + ) 1 1 13. = We know from trigonometry that, for any x and 2 = 2 any integer k , sin( x + 2 k π ) = sin( ) x . Since

1 10 sin( ) x =

where k is any integer.

Instructor’s Resource Manual

Review and Preview 345

14. We know from trigonometry that, for any x and

21. y ' = xy 2 → dy = xy dx 2

any integer k , cos( x + 2 k π ) = cos( ) x . Since

1 cos( ) π= -1, cos( ) x =− 1 2 dy = y xdx

if x =+ π 2 k π = (2 k + 1) π where k is any integer.

∫ dy

2 = y xdx ∫

15. We know from trigonometry that, for any x and

any integer k , tan(

π ) tan( ) . Since

−= x 2 + C

tan ⎜⎟ ⎛⎞= 1, tan( ) 1 if x = x =+ k π =

When x = and 0 y = 1 we get C = − . Thus, 1

⎝⎠ 4 4 4 2 where k is any integer. 1 1 2 x − −= 2 x −= 1

Since

16. sec( ) x =

, sec( ) is never 0 .

y =−

cos( ) x

17. In the triangle, relative to θ ,

1, hypot = so that x

tan θ = x 2 − 1 ydy ∫ cos x dx

When x = and 0 y = 4 we get C = 8 . Thus,

18. In the triangle, relative to θ ,

y 2 = sin x + 8

opp = x , adj = 1 − x 2 , hypot = so that 1 2 y x 2 = 2sin x + 16 sin θ = x cos θ = 1 − x 2 tan θ =

19. In the triangle, relative to θ ,

opp = 1, adj = x , hypot = 1 + x 2 so that

20. In the triangle, relative to θ ,

opp = 1 − x 2 , adj = x , hypot = 1 so that 1 − x 2

Review and Preview Instructor’s Resource Manual

Transcendental

6 Functions