Calculus Chapter 4 Solutions

Calculus Chapter 4 Solutions
1. a. The derivative or rate of the distance function is the velocity function. The area under
the velocity function is the reverse of the derivative: it gives the position function.
b. By taking the limit of slopes of smaller and smaller secants.
c. As we use more rectangles with smaller widths, we can try to take a limit of the area
approximations, which they give us.
2. a. We could take the limit of the approximations of area as the number of rectangles
goes to infinity.
b. As Δx gets smaller, potentially approaching 0.
3. lim
n!"
n#1
b#a
( )
$ f a + b#a
n n
i (using left rectangles)
i=0
a. No, we cannot have rectangles with a width of 0.
b. Each individual rectangle’s area would approach 0.
c. This is because the number of rectangles is increasing toward infinity, so the total
area need not approach 0.
4. a. The numbers b and a represent the right and left boundaries for x, respectively.
b. Instead of Δx, we now have dx in the integral.
c. f (x) represents the heights of all rectangles and dx represents the bases.
5. a. The region covers two cycles of the sine curve
(compressed by a factor of two), and it has equal
area above and below the x-axis:
2!
" sin(2t)dt = 0
0
b. See graph below left. The region is a trapezoid with area 2+4.5 ! 5 = 16.25
2
c. See graph below center. The region is a 4 X 6 rectangle with area 24.
Solution continues on next page. →
Chapter 4 Solutions
Calculus

5. Solution continued from previous page.
d. See graph below right. The region has width 0 and has area 0.
b. c. d.
6. These are areas of trapezoids:
10+25
a. !1 = 17.5 miles
10+70
b. ! 4 = 160 miles
2 2
10+17.5
c. ! 1 2 2 = 6.875 miles d. 10+10+15t
! t = (10 + 7.5t)t = 10t + 7.5t
2 2 miles
7. a. Even
2
2
b. " f (x)dx will be twice f (x)dx
!2
! , i.e., it is 20, because the area from x = !2 to 0
0
must equal the area from 0 to 2 due to symmetry.
c. This area must be the difference of the other two: !2 ! 10 = !12 .
2
d. We can halve " f (x)dx to get ! f (x)dx , and then we have:
3
!2
3
! f (x)dx = ! f (x)dx " 1 f (x)dx
2 !
2
0
2
"2
2
0
8. a. ! 4 5 x!3 ! 4 b. !x !1/2
c. !6 sin x d. f (x) = 4(x2 +1)
x 2 +1
= 4,! f !(x) = 0
15
9. a. # 1! f ("8 + i) = # 2("8 + i) 1/3 + 1 = 12
i=0
63
)
b.
1
4 ! f "8 + i 4
i=0
15
i=0
63
( ) =
1
)
i=0
( ) 1/3 + 1
#
2 "8 + i 4
$% 4
&
'( = 15
10. f (x) = x 2 + x ! 12,! f "(x) = 2x + 1 , f (3) = 0, f !(3) = 7 , tangent line: y = 7(x ! 3)
Chapter 4 Solutions
Calculus

n!1
11.
3
" 2 + 3 n i 3
= "
n f
i=0
( )
n!1
n
i=0
( 3 ( 2 + 3 n i ) + 5)
Larger n-values give better estimates, e.g., n = 30 yields 46.05. Using midpoint rectangles
would give the exact area of 46.5 for any n.
12. He is wrong because the triangle farthest from the pole sweeps out more volume when it is
rotated.
19
( )
3!
3!
13. a. 3 and ( sin # 2 "! +
40 $ 40 i %
& ' 2.94 , fairly close
i=0
19
( ) 2 ! 3
b. 53 1 3 and 1 "
' !5 + i 2 # 2
i=0
$
% & 53.75 , fairly close
14. See graph at right. Note: y-axis is scaled by 10’s.
They are next to each other and form (9x ! 2)dx = 250
"
8
!2
15. See graph at right. Note: y-axis is scaled by 5’s.
a. Geometrically it looks like it would have area 50,
but the calculator gives the (correct) answer of –50.
b. The bounds have been switched.
16. They will not be equal (unless they are both zero): they are opposites, as seen in the
limiting Riemann sum, in which b!a becomes !
b!a
= a!b if we were to switch a and b.
n n n
b
So " f (x)dx = !"
f (x)dx .
a
a
b
3
17. a. ! f (x)dx
b. ! f (x)dx
1
d
10
9
x+h
c. ! f (x)dx
d. ! f (t)dt
e
b
c
Sample conjecture: ! f (x)dx + ! f (x)dx = ! f (x)dx (which is basically part (c)).
a
b
c
a
x
18. All equal –4.5.
Chapter 4 Solutions
Calculus

19. a. F(x) = !2x + C b. F(x) = 3x 1/2 + C
c. F(x) = !x 3 + 3x 2 + C d. F(x) = (x + 3) 2 + C
20. a. f !(x) = 18(x " 2) 3
b. f !(x) = 2 cos x
c. f (x) = 2x 2 + 9x ! 5!"! f #(x) = 4x + 9
d. f (x) = x 2 ! 6x + 2!"! f #(x) = 2x ! 6 for x $ 0
21. A slope function returns the slope of the tangent to f(x) for any value of x, i.e., the exact
slope of f(x).
!
!
22. a. " x sin x dx = ! , 2 x sin x dx = 2!
0 " , x sin x = 2!
0 #"!
(because x sin x is an even function)
!
23. a. b.
24. a.
x
lim
2 "4
x!2
x"2 = lim (x+2)(x"2)
= lim (x + 2) = 4
x!2
x"2 x!2
(1+h)
b. lim
2 "1
= lim
1+2h+h 2 "1
= lim (2 + h) = 2
h!0
h h!0
h h!0
25. Similar triangles help us get the radius: r = 10!h
10 " 4 = 2 (10 ! h)
5
A = !r 2 = ! # $
2 (10 " h) %
5 &
2
=
4!
25
(10 " h)2
26. The rectangles still provide a good estimate of area.
Chapter 4 Solutions
Calculus

27. a. A = 0 because the width is zero.
b. The regions may be combined: ! f (x)dx + ! f (x)dx = ! f (x)dx
c. We are subtracting this area from itself: ! f (x)dx + f (x)dx
b ! = 0
a
d. We are amplifying the function by k, which multiplies the area
by k:
b
! kf (x)dx = k " ! f (x)dx
a
b
a
b
a
a
c
b
b
c
a
2
28. a. ! xdx + ! 3xdx = 1 2 x2 2 + 3 0 2 x2 2 (2 " 0) + (6 " 0) = 8
0
0
2
2
0
b. ! (4x)dx = 2x 2 2 = 8 " 0 = 0
0
!
0
b
c. ( f (x) + g(x))dx
a
29. a. No, adding k to f(x) shifts the whole region up, so the total area will be
b
! f (x)dx + (b " a)k .
a
b. No, these are different parts of the curve.
b
b+c
c. ! f (x)dx = f (x " c)dx
a ! is true because it shifts the curve and shifts the integral
a+c
so as to take the same area.
d. No, the second integral shifts the curve, so it takes a different area.
30. • ! f (x)dx = 0
a
a
b
• ! f (x)dx + ! f (x)dx = ! f (x)dx
a
b
c
b
b
• ! f (x)dx + ! g(x)dx = ! ( f (x) + g(x))dx
a
b
• ! f (x) dx = " ! f (x) dx
a
a
a
a
b
c
a
b
• " k ! f (x) dx = k ! " f (x) dx
b
b
• ! f (x)dx = ! f (x " h)dx
a
b+h
a+h
a
b
a
31. a. y = 1+ x !1 , y ! = "x "2 b. y ! = " sin x + cos x
c. y = x 5/3 , ! y ! = 5 3 x2/3 d. y = 6 ! 17x + 10x 2 , y " = !17 + 20x
Chapter 4 Solutions
Calculus

32. a. A trapezoid, A = 16+72 ! 7 = 308
2
b. The curve is translated up by 5, A = 308 + 5 ! 7 = 343
c. A rectangle, A = 5 ! 7 = 35
The area in (b) is the sum of those in (a) and (c), because its function is the sum of the
function in (a) and (c).
3+7
33. a. ! 2 = 10 b. 10
2
c. They are equivalent expressions; they just use different dummy variables.
34. a. f !(x) = 3x 2 " 2x + 1 = 2, 3x 2 " 2x " 1 = 0 , (3x + 1)(x ! 1) = 0 , x = ! 1 3 , 1;
( ) = ! 1 27 ! 1 9 ! 1 3 + 1 = 14
27
( ) or y = 2x + 32
29
( ) + 14
f ! 1 3
The lines are y ! 14
27 = 2 x + 1 3
b. y = ! 1 2 x + 1 3
, f (1) = 1 ! 1+ 1 + 1 = 2 ,
27 and y = ! 1 2 (x ! 1) + 2 = ! 1 2 x + 5 2
, and y ! 2 = 2(x ! 1) or y = 2x
35. a. f (g(x)) b. g( f (x))
c. h( f (x)) d. g(h( f (h(x))))
36. a. Negative b. Zero c. Positive
d. Zero e. Negative
37. The secant line goes through (0, 0) and (2, 0) and has slope 0, so we
set f !(x) = 0!" 3x 2 ! 4x = 0
x(3x ! 4) = 0!"!x = 0,! 4 3
f (0) = 0 , so at (0, 0) and
( ) = ( 4 3 ) 3 ! 2 ( 4 3 ) 2 = 64
f
4
3
27 ! 32
9 " 3 3 = ! 32
27 , so at 4
( 3 , ! 32
27 ).
38. See graph at right. y-axis is scaled by 30’s, x-axis is scaled by 2’s.
a. f !(x) = 3x 2 + 6x " 45
f !("2) = 3("2) 2 + 6("2) " 45 = 12 " 12 " 45 = "45
b. This is where !! f (x) = 6x + 6 = 0 , x = !1, f (x) = 55 ;
The inflection point is (–1, 55).
Chapter 4 Solutions
Calculus

39. a. !2(3) ! 5 = !11 b. 2(3) 2 ! 4 = 14
c. f(x) is discontinuous and non-differentiable at x = 3 and lim f (x) does not exist.
x!3
40. a. 0 b. 5 c. 10
d. 5x e. cx
41. a. (trapezoid) 5+13 ! 2 = 18
5+41
b. ! 2 = 207
2 2
5+(4 x+5)
c.
! x = (2x + 5)x = 2x
2 2 + 5x
d. The area under the curve for 0 ! t ! x .
e. A(x) = m 2 x2 + bx ; mt + b is the equation of a line of slope m through (0, b) .
42. b. B(x) = 13+4x+5 !(x " 2) = (2x + 9)(x " 2) = 2x
2 2 + 9x " 4x " 18 = 2x 2 + 5x " 18 , and it
differs from A(x) by a constant.
c. B(x) = (2x 2 + 5x) ! 18 = A(x) ! A(2) .
x
These two coincide to the areas ! (4t + 5)dt = (4t + 5)dt " (4t + 5)dt
2 ! 0 ! .
0
d. Yes, they grow at the same rate because any incremental change in x (Δx) adds and
equal amount of area to each.
e. G(x) = A(x) ! A(c) is the area under f(x) from c to x.
x
2
43. We can get distance from velocity by taking the area function (for area under the velocity
curve), and we can get velocity from distance by taking the slope function of the distance.
44. a. sin x + C b. !2 " x!1 +C
!1
= 2x !1 + C
c. !9 " x4/5
4/3 + C = ! 27 4 x4/3 + C
19
( ) 3/2
45. a.
1 "
2 ! 4 i $
' 5 # 5 % & !40. 04 , integral is –43.2
i=0
19
b.
7i
( ) + 3
! 4 1 + 7i " 30.85 , integral is 31.42
20 20
i=0
46. Any constant has a derivative of zero, so any constant can be added to one antiderivative
without changing its derivative, yielding infinite possibilities for the antiderivative.
Chapter 4 Solutions
Calculus

47. f (x) = x 3 ! 4x 2 + x ! 4 , f !(x) = 3x 2 " 8x + 1, f (!3) = !27 ! 36 ! 3 ! 4 = !70 ,
f !(3) = 27 + 24 + 1 = 52 , tangent line is y + 70 = 52(x + 3) or y = 52x + 86 .
48. No, for example for problem 4-47, d dx (x2 + 1) = 2x and d (x ! 4) = 1 and 2x !1 = 2x is
dx
not the derivative.
49. a. f (x) = x and g(x) = sin(x 2 ) + 1
b. f (x) = x 2 + 2 and g(x) = 3x 3 ! 12
50. Sample answer:
1
x dx
!
1
0
6
51. ! 5dx = 5x 6 = 30 15 = 5dx
0
0 0
15 = 5x
3 = x
No, this is not the same line that will divide
!
x
6
! 5xdx = 5 2 (6)2 = 90 in half.
0
3
52. ! f (t)dt = ! f (t)dt " ! f (t)dt = 9(3) 2 " 2 " [9(2) 2 " 2] = 79 " 34 = 45
2
3
0
2
0
53. a. Definite b. Indefinite c. Definite
d. Indefinite e. Indefinite
54. a. A(x) = 2(x ! c) = 2x ! 2c = 2x + C
b. A(x) = (5c+2)+(5x+2)
2
!(x " c)
= 1 2
c. A(x) = 3 2!c
(5x + 5c + 4)(x " c)
= 5 2 x2 + 2x " 5 2 c2 " 2c
= 5 2 x2 + 2x + C
( )+ 3 2!x
( ) "(x ! c) = 1 (3 ! c ! x)(x ! c)
2 2
= ! 1 2 x2 + 3 2 x ! 3 2 c + 1 2 c2 = ! 1 2 x2 + 3 2 x + C
Chapter 4 Solutions
Calculus

55. a. 2 ! 8 " 2 ! 3 = 10
b. 5 ! 8 " ("2) ! 8 = 56
( ) = 3!(102 " 14) = 264
c. 3 5 2 ! 62 + 2 ! 6 " ( 5 2 ! 22 + 2 ! 2)
#
d. (5x + 2)dx = 5 2 ! 32 + 2 ! 3 "
1
3
10
( ) dk
3
e. " 2 ! k = 3 2 k ! 1 2 k2 10 = 30
4
4
!2
( ) dk
3
f. " 2 ! k = 3 2 k ! 1 2
0
( ) = 28.5 " 4.5 = 24
5
2 !12 + 2 !1
( 2 ! 100
2 ) ! 12 2 ! 16 2
!2
k2 = !6
0 2 ! 4 2
( ) ! (0) = !5
( ) = !35 ! (!2) = !33
3
56. a. ! g(m)dm = 4(3)+1
0
0
3+2
= 13 5
b. " g(m)dm = !" g(m)dm = 4(!1)+1 = ! !3
!1+2 1 = 3
!1
5
!1
c. " g(m)dm = 4(5)+1
!1
0
5+2 ! 4(!1)+1 = 21
!1+2 7 ! !3
1 = 6
57. Ignacio is correct.
58. a. d !(t) = sin t , d !(1) = sin1 " 0.84 , d !(" ) = 0 , d !(50) = sin 5 " #0.96
This uses the derivative as the velocity, particularly the derivative of
cos t , and also the derivative of a multiple (in this case of –1) of a
function and the derivative of a sum of two functions.
b. See graph at right. Area = 1 2
! 5 !15 =
75
2
= 37.5 ft
c. Velocity is the derivative of distance and distance is the area under the velocity curve.
59. a. f (1) = 50 miles
b.
50 miles
1 hour
= 50 mph
c. 75 mph, at 0.5 hours (30 minutes)
Chapter 4 Solutions
Calculus

#
1
60. (x 1/3 ! 4)dx " !13.14
!2
61. Yes, f must be continuous at x = c to be differentiable there, so f (c) must exist.
!
62. f !(x) = cos x , f
tangent line is y ! 3
2 = 1 2 x ! " 3
( 3 ) = sin 3
2 , f ! "
( 3 ) = cos " 3 = 1 2 ,
( ) or y = 1 2 x + 3
2 ! " 6
63. a.
7
3 x2/3 b. !21m !8 ! 21m 2
c. 0 d.
d
dt (6t 2 + 15t) = 12t + 15
64. a. 3x 5 + 2x 2 ! 3x + C b. !2 sin x + C
c. !x 4 + 10x + C
65. i. F(x) = 2x + C , A(x) = 2x
ii. F(x) = x 2 + C , A(x) = x 2
A(x) + C = F(x)
"
! /3
66. a. cos x dx
! /6
b. Answers vary, you may suggest Tommy use the antiderivative F(x) = ! sin x + C .
c. F(x) = ! sin x + C
( ) + C) ! (! sin ( " 6 ) + C) , ! 3
d. (! sin " 3
2 + 1 2 = 1! 3 un
2 2
e. The +C is added and subtracted when evaluating, therefore it is eliminated.
67. a. ! (2x + 5)dx = x 2 + 5x
b. A !(x) = f (x) ; Geometrically, f (x) is the slope function of area, A(x) .
c. A(x) = 1 2 mx2 + bx + C and A !(x) = mx + b
Chapter 4 Solutions
Calculus

68. a. As x increases, there is more area under the curve.
! f (x)dx = A(x)!!"!! d f (x)dx
dx ! = d A(x)!!"!! f (x) = #
dx A (x)
b. Since F(c) is a constant, this expression becomes f (x) + 0 = f (x) .
c. (1) Since f (x) is continuous, the anti derivative exists, or F(x) = ! f (t)dt .
a
(2) F(x) means geometrically, the area under the curve from a to x.
(3) Using (1), take the derivative of both sides: d dx F(x) = d f (t)dt
dx ! .
(4)
F(x+h)"F(x)
Rewrite the left side using the definition of the derivative: lim
h!0
h
(5) The definition of the derivative for F(x): F(x + h) ! F(x) " hf (x) .
(6) Divide by h and apply the limit as h ! 0 . The derivative becomes f(x).
(7) Simplify your results: f (x) = d f (t)dt
dx ! .
a
d. " f !(x)dx
=
d
" dx f (x)dx = f (x) + C
x
x
a
x
69. A(x) = F(x) + C
"
70. F(x) = 2x 3 + 7x + C , (6x 2 + 7)dx = F(2) ! F(1) = (16 + 14 + C) ! (2 + 7 + C) = 21
1
2
#
x
71. a. F(x) = x 2 + 5 + C , (2x + 5)dx = x 2 ! 5x + C ! (0 2 + 5 " 0 + C) = x 2 + 5x
#
9
0
b. (2x + 5)dx = F(9) ! F(4) = (9 2 + 5 " 9) ! (4 2 + 5 " 4) = 81 + 45 ! 36 = 90
4
72. F(x) = x 2 + 5x + C ,
9
! (2x + 5)dx = (9 2 + 5(9) + C) " (4 2 + 5(4) + C) = (81+ 45) " (16 + 20) = 90
4
"
2
73. a. (6x 3 + 9x)dx
!5
9
b. ! h(x) dx " ! h(x) dx = "! h(x)dx!!or!! ! h(x)dx
#
5
9
3
c. ![(2x) 2 + ( x ) 2 ]dx = ! "(4x 2 + x)dx
!
7
2
#
7
2
d. (x + 5) 2 dx + xdx = x 2 + xdx = (x 2 + x )dx
1
4
(by translating the first integral)
!
9
6
!
9
6
5
3
!
9
6
3
5
!
9
6
Chapter 4 Solutions
Calculus

74. ! f (t)dt represents total weight loss, in pounds, between day a and day b.
a
75. a. ∞ or does not exist b. does not exist
c. 2 d. –2
e. 2 f. 0
g. At x = –2, –1, and 2 (and below –2)
76. a. 17 1 3
b. The area under v(t) from t = 0 to 4, and this is the distance at t = 4 , i.e. s(4) .
c. acceleration
d. feet
2
77. a. " ! f (x)dx = " f (x)dx = 10
4
12
4
2
b. " ( f (x ! 8) + 4) dx = " f (x)dx + " 4dx = 10 + 4x 12 = 10 + (48 ! 40) = 18
10
12
c. ! f (x + 8)dx = ! f (x); not enough info
10
20
18
4
2
10
10
78. ! 2xdx = x 2 10 = 100 Solve: 0 x2 x = 0 50!!!!!x2 = 50!!!!!x = 50 = 5 2
0
79. a. s !(t) = 4 " 1 2 t #1/2 = 2t #1/2 , s !(1) = 2 , s !(4) = 2 = 1, s !(16)
= 2 = 1 4 16 2
This uses derivative to obtain velocity, the Power Rule for this particular derivative,
and the rule d dx (c ! f (x)) = c ! d dx ( f (x)) .
1
1
b. s(1) = ! v(t)dt = ! (9t + 32)dt , ! v(t)dt = 9 2 t 2 + 32t + C ,
&
1
0
0
(9t + 32)dt = 9 2 (1)2 + 32(1) ! "
#
9 2 (0)2 + 32(0) $
% = 36.5 miles
0
This uses a definite integral and the fundamental theorem to use the antiderivative to
calculate this integral.
c. Distance is an antiderivative or area function of velocity, and velocity is the
derivative or slope function of area.
Chapter 4 Solutions
Calculus

4-80. a. See graph at right.
!3
i. g(!3) = " f (t)dt = !"
f (t)dt = ! 1 2 # !3# 3
0
!7
0
!3
0
( ) = 4.5
( )
ii. 2g(!7) = 2" f (t)dt = !2 f (t)dt
0 " = !2 1 (!4)(4) + (!4)(3)
!7
2
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! = !2(!20) = !40
9
iii. g(9) + g(!9) = " f (t)dt ! f (t)dt
0 "!9
= ( 1 (4)(4) + 4(5)
2 ) ! 1 (4)(!4) + (!4)(5)
2
b. See graph at right.
!3
0
( ) = 28 ! (!28) = 56
i. g(!3) = " f (t)dt = !"
f (t)dt = ! 1 2 # 3# 3
0
!7
0
!3
( ) = !4.5
ii. 2g(!7) = 2" f (t)dt = !2 f (t)dt
0 " = !2 1 (4)(4) + (4)(3)
!7
2
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! = !2(20) = !40
iii. g(9) + g(!9) = " f (t)dt ! f (t)dt
0 "!9
= ( 1 (4)(4) + 4(5)
2 ) ! 1 (4)(4) + (4)(5)
2
9
0
0
( )
( ) = 28 ! 28 = 0
y
y
x
x
82. a. The areas under the curve from 0 to a and from 0 to b, respectively.
b
b. ! f (x)dx = ! f (x)dx "!
f (x)dx = F(b) " F(a)
a
t
b
0
c. F !(t) = d (2t + 3)
dt " dt = d dt (t 2 + 3t # 0) = 2t + 3
0
a
0
2
83. a. 2 hours: s(2) = ! (2t + 5)dt = t 2 + 5t 2 = 0 22 + 5(2) " (0 + 0) = 14 miles
0
5
5 hours: s(5) = ! (2t + 5)dt = t 2 + 5t 5 = 0 52 + 5(5) " (0 + 0) = 50 miles
0
x
x hours: s(x) = ! (2t + 5)dt = t 2 + 5t x = 0 x2 + 5x
4
0
b. s(4) ! s(2) = " (2t + 5)dt = t 2 + 5t 4 = 2 42 + 5(4) ! (2 2 + 5(2)) = 22 miles
2
c. The derivative is the slope of the tangent line and slope is the rate of change. The rate
of change involving distance is velocity.
Chapter 4 Solutions
Calculus

84. a. 3x ! 5
x
d
b. " (3x ! 5)dx = 3dx
dx " = 3x x = 3x ! 9
3
3
x
3
c. 0; The integration will yield a constant. The derivative of a constant is 0.
d.
d
" (3x ! 5)dx = 3
dx " dx = 3x + C
e. cos(x 2 )
d
f. ! dx (cos x2 )dx = cos x 2 4 = cos(16) " cos(1)
1
1
4
85. Note: The FTC gives exact results, whereas calculators give very close estimates.
a. 2 !
x "2
"2
8
1
8
= "x "2 1
= " 1
63
" ("1) =
64 64
" /2
b. ! cos x " /4 = ! cos
"
2 ! (! cos " 4 ) = 0 + 2
2 = 2
2
c. 3 x3/2
9 9
3/2 = 2x
3/2 = 2 ! 27 " 2 ! 8 = 54 " 16 = 38
4 4
d. x 3 ! 3x 2 + 2x 0
2
= 8 ! 3" 4 + 2 " 2 ! 0 = 8 ! 12 + 4 = 0
#
e. 2 ! 3! x"1
$
%
"1 1
3
&
'
( = "6x"1 3
= "
6
1 3 " ("6) = 4
x
f. 4/3 8
4/3 =
3
4 x4/3 8
=
3
!1
!1 4 "16 ! 3 (1) =
45
4 4
9
g. Combine: (2x ! 20)dx = x 2 9
! 20x = (81! 180) ! (4 + 40) = !99 ! 44 = !143
"!2
!2
( )
h. 9 m1/3
1/3 ! 2 m!2
27
!2 = 27m 1/3 + m !2 27
= 81 +
1
1 1
27 2
! (27 + 1) " 53.001
86. He did not distribute the negative. So it should have ended in line 2, with –2 – C. Then the
C’s would subtract out giving 68.
Chapter 4 Solutions
Calculus

87. a.
3
2 x4 ! x 2 + 5x + C
b.
3
2 x4 ! x 2 4
+ 5x =
3
" 256 ! 16 + 20
2 2
c. 3t 3 ! t + C
d. 3t 3 ! t !2
e. ! cos m + 1 9 m3 + C
( ) ! ( 3 "16 ! 4 + 10
2 ) = 384 + 4 ! 30 = 358
2
= (3" 8 ! 2) ! (3(!8) ! (!2)) = 22 ! (!22) = 44
f. ! cos m + 1 9 m3 = ! cos " + 1 !"
9 " 3
"
( ) ! (! cos(!" ) + 1 (!" )3
9 ) = 1+ " 3
9 + (!1) + " 3
9 = 2" 3
9
!3
88. a. " ! f (x)dx + " g(x)dx = " (g(x) ! f (x))dx
!
!5
!3
!5
b. (3 f (x) + 5g(x))dx
1
6
11
c. ! f (x)dx or ! f (x)dx or just zero.
!
11
10
d. f (t)dt
9
6
6
!3
!5
f (x+h)# f (x#h) 4(x+h)
89. f !(x) = lim
= lim
2 #9(x+h)#1#[4(x#h) 2 #9(x#h)#1]
=
h"0
2h h"0
2h
4 x
lim
2 +8xh+4h 2 "9x"9h"1"4 x 2 +8xh"4h 2 +9x"9h+1 16xh"18h
= lim = lim 8x " 9 = 8x " 9
h!0
2h
h!0
2h h!0
90. a. cos(x ! 5) b. 82x 81
c. !2 3 2
( ) m 1/2 = !3m 1/2 d. 3(x + 6) 2
91. f !(x) = 3x 2 + 24x + 36 = 3(x + 2)(x + 6), and f !!(x) = 6x + 24 . From f !(x) , we know that
f (x) is increasing in (–∞, –6] and [–2, ∞) (where f !(x) " 0 ), and f (x) is decreasing in
[–6, –2] ( f !(x) " 0 ). From f !!(x) , the point of inflection is at x = !4 and is (!4, !22),
and f is concave down in (–∞, –4] and concave up in [–4, ∞).
92. Its inflection point (from #78) was (–4, –22), where the slope is
f !("4) = 3("4) 2 + 24("4) + 36 = "12 .
The tangent line is y + 22 = !12(x + 4) or y = !12x ! 70 .
Chapter 4 Solutions
Calculus

93. a. The tangent is vertical; the slope goes to infinity and no actual numerical derivative
exists.
b. No, the tangent is still vertical (and in this case there is a cusp).
c. This is just f (x) = x and it is differentiable everywhere.
d. Because the graph changes from concave up to concave down.
2
94. ! (g(x) + 3)dx = 0 (because of the bounds)
2
2
" (3 ! 5g(x))dx = !3x 4 + 5 g(x)
2 " = !3(4) + 3(2) + 5(6) = 24
4
4
2
96. The defendant traveled 12 + 15 + 21+ 26 + 10 + 16 = 100 miles in 2 ! 18 = 2.3 hours, at an
60
average velocity of 43.48 mph, so the officer is incorrect, the teacher may be innocent. But
this does not prove innocence.
97. This is at the inflection point s !(t) = "93.75 # 3t 2 + 93.75 # 2t + 45 ,
s !!(t) = "93.75 # 6t + 93.75 # 2 = 0 , t = 93.75!2
93.75!6 = 1 3
1
s!
3
hour , and this velocity is
( ) = "93.75 # 3( 1 3 ) 2 + 93.75 # 2 ( 1 3 ) + 45 = 76.25 mph, which confirms the prosecution’s
claim.
$
98. a. cos x is even, so cos xdx = 2 ! cos xdx = 2 !sin x 0
" = 2 !(0 # 0) = 0
b. 5 ! y3/2
3/2 + cos y + C = 10 3 y3/2 + cos y + C
"
#"
$
"
0
c. The antiderivative of 1 x
has not yet been taught, but with a calculator this is ! 1.
d. ! 1dx = x + C
99. a. The graph represents distance and s !(t)is velocity.
b. The graph represents sales rate (in sales/hr) and ! r(t)dt represents and indefinite
area of total sales.
c. The graph represents a weight rate in lbs/inch, and w(i)di ! represents an indefinite
area or lbs.
Chapter 4 Solutions
Calculus

#%
100. a. v(t) = ! 20 t + 20, 0 " t " 4.5
3
$
&% !!20t ! 100, 4.5 < t " 5
b. For 3 hours, the triangle has area 1 2 ! 20 ! 3 = 30 miles. For the 4th hour we should
count the area as positive, 1 2 ! 20 3 !1 = 10 3 , for a total of 33 1 3 miles.
c. The car was traveling south, towards its starting point.
!
5
d. v(t)dt
!
0
5
e. v(t) dt
0
101. a. The backwards/forwards nature of finding velocity from position and vice versa is
accomplished using the derivative and integral, as proscribed in the Fundamental
Theorem.
b. We must consider integrating both parts of the function:
#%
s(t) = ! 10 3 t 2 + 20t, 0 " t " 4.5
$
&% 22.5 + 10t 2 ! 100t + 247.5 = 10t 2 ! 100t + 270, 4.5 < t " 5
The 22.5 is ! 10 3 t 2 + 20t evaluated at t = 4.5 , and 247.5 is the opposite of
10t 2 ! 100t evaluated at 4.5, in order to make s(t) continuous at 4.5.
102. y ! = 3x 2 " 14x + 15 = (3x " 5)(x " 3) is non-negative in (!", 5 3 ) ! [3, ") , which is where y
5
is increasing. It is non-positive in !
" 3 , 3 #
$ , where y is decreasing.
103. a. y ! = cos(x " 3) , differentiable everywhere
#
b. f !(x) = "2x, x < 1
$
3(x " 1) 2 , it is not differentiable at x = 1 because the limits (of
%& , x > 1
secants) are different from the left and right sides.
%
2
104. 6 x dx !
2
0
n#1
$
n " 621/n
i=0
Chapter 4 Solutions
Calculus

105. a. f (x) = 2x 3 + 6x 2 ! 7x + C
b. 25 = f (!2)
25 = 2(!2) 3 + 6(!2) 2 ! 7(!2) + C
25 = !16 + 24 + 14 + C
C = 3
Therefore f (x) = 2x 3 + 6x 2 ! 7x + 3
c. The integral of the derivative gives the original function plus a constant. Using the
point provided, we can solve for the constant.
106. a. The velocity is negative in the second interval, and s(1.6) = 84 and s(1.8) ! 74 .
b. Look at the inflection point in each section:
1 st section: s !(t) = "291.67 # 3t 2 + 1125 # 2t " 1360 ,
s !!(t) = "291.67 # 6t + 1125 # 2 = 0 , t = 2!425
291.67!6 " 1.286 , s ! (1.286) " 86.4 mph
2 nd section: s !(t) = 2500 " 3t 2 # 12750 " 2t + 21600 , s !!(t) = 2500 " 6t # 12750 " 2 = 0 ,
t = 12750!2
12750!6 = 1.7 , s ! (1.7) = "75 mph so max speed is 75 mph
3 rd section: s !(t) = "176.3t 2 + 1142.4 # 2t " 2402 , s !!(t) = "176 # 6t + 1142.4 # 2 = 0 ,
t = 1142.4!2 " 2.164 , s !(2.164) " 69.7 mph
176!6
Basically the inspector is right, there are two sections where 70 mph is exceeded.
107. Exhibit B does not meet the starting point of Exhibit C, portraying a discontinuity. At time
t = 1hour, Exhibit C shows the teacher at a position of 48 miles, whereas the function of
Exhibit B shows a position 45 miles. Exhibit B shows that the teacher arrived in D’exdete
before an hour. (But if the teacher stopped in D’exdete, she should have left at an initial
velocity of 0 for her distance function to be differentiable, but Exhibit C’s function starts
with a derivative of 15mph).
108. a. " (2y ! 3) 2 dy = " (4y 2 ! 12y + 9)dy = 4 3 y3 ! 6y 2 + 9y + C ; Expanded the function
and then the reverse Power Rule was used for each term.
b. ! 26.93 , using a calculator
c. 9 ! m"1
"1 + 7 ! m12
12 + C = "9m"1 + 7
12 m12 + C ; The reverse Power Rule was used for
each term.
d. ! 7.069 , using a calculator
Chapter 4 Solutions
Calculus

109. y = x(x 2 + 2x ! 3) = x(x ! 1)(x + 3) crosses the x-axis at x = !3, so this is
0
(x 3 + 2x 2 ! 3x)dx = x4 4 + 2x3
3 ! 3x2
0
(!3)
" 2
2 = ! ! 2(!3)3 + 3(!3)2 = 11.25un 4 3 2 2
!3
!3
"
9
110. a. [ f (x) 2 ! g(x) 2 ]dx
!
2
9
b. f (x)dx
3
1
c. " f (x)dx + " f (x)dx + " f (x)dx = " f (x)dx
!
!2
8
d. (2k(x) + j(x))dx
2
1
4
9
4
9
!2
#(x+h)
111. a. f !(x) = lim
2 +3(x+h)+1#(#x 2 +3x+1)
= lim
#2xh#h 2 +3h
= lim # 2x # h + 3 = #2x + 3
h"0
h
h"0
h h"0
b. f !(x) = "2x + 3 , the same
c. f !(0) = 3, f !(1) = 1
112. a. When f !(x) is non-negative: (!", !5] ! [1, 6]
b. When f !(x) is increasing, roughly [–2.5, 3.7].
c. Negative, because at x = 8 the slope of f '(x) is negative and f !! is the slope
function of f ! .
113. a. It is a cylinder with cones carved out of its top and bottom.
b. !r 2 h " 2 (!r 2 # 4 # 1 3 ) = ! # 25 # 8 " 8 3 ! # 25 = 2 400!
(! # 200) = un
3 3 3
19
114.
1
# 5 (!4 + i 5 ) " 2!2+i/5 $ !6.448
i=0
3
115. ! (h(x) + j(x))dx = "! (h(x) + j(x))dx = "! h(x)dx " ! j(x)dx = " (4 + 2) = "6
5
7
5
3
5
" (h(x ! 2) + 2)dx = " h(x)dx + " 2dx = 4 + 2x 7 = 2 + (14 ! 10) = 8
5
5
3
7
5
5
3
5
3
Chapter 4 Solutions
Calculus

116. All of the velocities basically connect between the pieces. The total area ! v(t)dt can be
0
obtained by adding each area separately:
0.7
" (!187.5t 2 + 150t + 45)dt + " (43.75t + 27.5)dt + " (!500t 2 + 1200t ! 640)dt +
0
1.8
1.2
0.7
" (7500t 2 ! 25500t + 21600)dt + " (!528t 2 + 2284.8t ! 2402)dt =
1.6
#
!187.5 " t3 3 + 150 t2 2
$
%
+ 45t 0
0.7
23
1.8
1.6
1.2
&
'
( + "
43.75 ! t2 2 + 27.5t 1.2 %
#
$
0.7 &
' + "
!500 t3 3 + 1200 t2 2 ! 640t 1.6 %
#
$
1.2 &
' +
#
7500 ! t3 3 " 25500 ! t2 2 + 21600t 1.8 &
$
%
1.6 '
( + #
!528 " t3 3 + 2284.8 t2 2 ! 2402t 2.3 &
$
%
1.8 '
( )
46.81+ 34.53 + 21.33 + (!10) + 25.96 " 118.63 miles.
This represents the total displacement and should be 100 miles according to Exhibit A, and
represents a contradiction in the data.
2.3
4
117. a. " (!x 2 + 6x ! 3)dx = ! x3 1
3 + 6 # x2 2 ! 3x 4
= !
64
+ 3#16 ! 12
3
1
"
4
b. (x + 1)dx = x2 2 + x 1
1
4
=
16
2 + 4 ! ( 1 + 1) = 10.5
2
( ) ! ( 1 3 + 3 ! 3 ) = 15
c. " ( f (x) ! g(x))dx = " (!x 2 + 6x ! 3 ! x ! 1)dx = " (!x 2 + 5x ! 4)dx
1
4
1
4
= ! x3 3 + 5 # x2 2 ! 4x 4
= 2
2
3 ! (!1 5 6 ) = 4.5
1
This can also be seen as part (a) minus part (b).
1
4
118. a. Length ! f (x) " g(x), width = !x , area ! "x( f (x) # g(x))
"
b
b. ( f (x) ! g(x))dx
a
c. The x-values of the points where the curves intersect.
119. a. Length ! g(x) " f (x) = "x + 8 " ((x " 3) 2 " 1) = "x 2 + 5x , width = !x ,
area ! "x(#x 2 + 5x)
5
b. " (!x 2 + 5x)dx = ! x3 3 + 5x2
5
2 = !
125
3 + 125
2 = 125
6 = 20 5 6
0
2
2
120. a. " !x ! (x 2 ! 6)dx = 22 3 un2
0
b. Because we are finding the area between two curves. When g(x) is subtracted, the
negative area becomes positive.
Chapter 4 Solutions
Calculus

121. a. width = !x , length = !(x ! 3) 2 + 9 ! (x + 6) = !x 2 + 5x ! 6 = !(x ! 2)(x ! 3),
3
Area = # (!x 2 + 5x ! 6)dx = ! x3 3 + 5 " x2 2 ! 6x ! 27
2
3 + 45
2 ! 18 ! ! 8 3 + 20
2 ! 12
2
( ) = 1 6 un2
= ! 9 2 ! ! 14 3
b. width = !x , length = sin x ! (x 2 ! 1) ,
1.410
3
=
Area = " (sin x ! x 2 + 1)dx = ! cos x ! 1 3 x3 + x 1.410
!0.637
= 0.315 ! (!1.355) # 1.671un 2
!0.637
( )
!2
123. a. " (3m 2 + 2m ! 9)dm = m 3 + m 2 ! 9m
!5
!5
!!!!!= !8 + 4 + 18 ! (!125 + 25 + 45) = 14 ! (!55) = 69
2
b. " (2t 2 + 3t)dt = 2 t3 !1
3 + 3 t2 2 !1
!1
( ) dx
2
=
16
!2
( ) = 10.5
3 + 6 ! ! 2 3 + 3 2
c. " 1 + 2 x + 1 , of which the middle term requires a calculator: ! 0.977
!4
x 2
d. a x2 2 + bx 3
= a !
9
2 + 3b
2
( ) " ( a ! 4 2 + 2b ) = 5a
2 + b
124. From the fundamental Theorem, F(x) is an antiderivative of f(x), so f(x) must be the
derivative of F(x): f (x) = d dx (3(x ! 4)3 + 6) = 9(x ! 4) 2 .
125. a. A hollow sphere.
b. ! 4 3 "r 1 3 , ! 4 3 "r 2 3 = 4 3 " # 64 ! 4 3 " # 27 = 4 148"
" # 37 = un
3 3 3
126. Yes, because: (1) It is continuous at x = 2 (both functions approach 1).
(2) d dx ((x ! 1)2 ) = 2(x ! 1) approaches 2 and d 2 sin(x ! 2) = 2 cos(x ! 2) also approaches 2.
dx
127. a. Distance is an antiderivative: " (!32 ! 18)dt = !16t 2 ! 18t + C = s(t) and
500 = s(0) = !16(0) 2 ! 18(0) + C = C ,so s(t) = !16t 2 ! 18t + 500 .
So s(1) = !16(1) 2 ! 18(1) + 500 = 466 ft, s(2) = !16(2) 2 ! 18(2) + 500 = 400 ft,
and s(3) = !16(3) 2 ! 18(3) + 500 = !144 ! 54 + 500 = 302 ft.
b. s(t) = !16t 2 ! 18t + 500 is an antiderivative of v(t).
Chapter 4 Solutions
Calculus

128. a. f (x) = (x!3)(x+1)
x!2
= x ! 3
x!2 : As x ! "#, f (x) ! "# . As x ! 2" , f (x) ! # .
As x ! 2 + , f (x) ! 0 ; as x ! ", f (x) ! " . End behavior: y = x .
b. As x ! "#, f (x) ! 0 ; as x ! 0 " , f (x) ! "# ; as x ! 0 + , f (x) ! " ;
as x ! ", f (x) ! 0 .
129. No, d dx ((x ! 3)(2x ! 9)) = d dx (2x2 + 3x ! 27) = 4x + 3, but d dx (x ! 3) " d (2x + 9) = 1! 2 = 2 .
dx
130. a. lim
x!9
b. lim
h!0
2+h " 2
h
#
x "3
x"9 # x +3
= lim x"9
= lim
x +3 x!9 (x"9)( x+3) x!9
2 x +1
c. lim = lim
x!" 5# x x!"
1
x +3 = 1 6
2+h + 2
= lim 2+h"2
= lim
2+h + 2 h!0 h( 2+h + 2 ) h!0
2 x
x + 1 x
5
x # x x
= lim
2
x!"
#1 = #2
1
= 1 = 2
2+h + 2 2 2 4
d. Cos(x) oscillates between 1 and –1, therefore the limit does not exist.
131. a. Width = !x , length = !2(x 2 ! 1) ! (!x 2 + 1) = 1 ! x 2 ,
1
Area = " (1 ! x 2 )dx = x ! x3 !1
3 !1
1
=
2
b. Width = !x, length = sin x " 3 4 x + 7 ,
2"
( ) dx
( ) = 4 3 un2
3 ! ! 2 3
Area = # sin x ! 3 4 x + 7 = ! cos x ! 3 "
4 $ x2 2 + 7x "
( ) # 8.89 un 2
= !1 ! 3" 2
2
+ 14" ! 1! 3 8 " 2 + 7"
2"
132. This would give a negative area; it uses rectangles with negative height.
133. a. The upper bound has two parts, one of which is x 2 and the other is !x + 6 .
2
b. (x 2 4
#
" ! x )dx = (!x + 6 ! x )dx
1 " = x3 2
3 ! x3/2
$
%
3/2 1
2
&
'
( + "
! x2 + 6x !
x3/2
2
#
$
3/2 2
4
8
3 ! 2 2
3/2 ! (1 3 ! 2 3 ) + (! 16 2 + 24 ! 8
3/2 ) ! (! 4 2 + 12 ! 2 x
3/2 ) ! 1.114 + 2.552 ! 3 2 3 un2
%
&
' =
Chapter 4 Solutions
Calculus

134. a. The functions cross paths, so the first part requires an integral of g(x) ! f (x) and the
second requires an integral of f (x) ! g(x) .
1
b. " (x 3 ! 2x 2 + 1 ! (x ! 1))dx + " (x ! 1 ! (x 3 ! 2x 2 + 1))dx
!1
1
= " (x 3 ! 2x 2 ! x + 2)dx + " (!x 3 + 2x 2 + x ! 2)dx
!1
"
= x4 4 ! 2 x3 3 ! x2 2 + 2x 1 %
#
$
!1&
' + "
! x4 4 + 2 x3 3 + + x2 2 ! 2x 2 %
#
$
1 &
'
1
2
1
2
( 3.083 or 3
1
12
2
5
"
135. a. " (2 ! x + 3)dx + (x ! 2 + 3)dx
0 " = ! x2 2
2 + 5x 2 %
#
$
0 &
' + "
x2 2 + x 5 %
#
$
2 &
' =
!2 + 10 ! (0) + ( 25 + 5) ! (2 + 2) = 8 + 13.5 = 21.5 , by splitting the integral into two
2
parts so as to remove the absolute value.
b. " (4m !3 ! 3 cos m)dm = 4 m!2
!2 ! 3 sin m + C = !2m!2 ! 3sin m + C by
antidifferentiating each term.
c. ! 2.050 by calculator integration
d. ! 2 x + C , because ! 2 is just a constant.
136. The most evident solutions are symmetric across the line y = x , but other solutions do exist
(e.g. the rectangle with 0 ! x ! 3, 1 ! y ! 2 ).
137. a. 0
b. lim
x 2 #2x+1 1#4x+(1 x
x!" x 3 = lim
2 )
= 0
x!"
x
cos x+1 # cos(x#" )+1 #(cos(x#" )#1)
c. lim = lim
= lim
= #0 = 0
x!"
x#"
x!"
x#"
x!"
x#"
138. a. lim
2 x "2
x!1 3 x # 0.42 b. lim
1"cos x
"3 x!0 x 2 # 0.50
139. a. sin( x ! 2 2 ) = sin(x ! 2) for x ! 2
b. sin 1 x ! 2
c.
1
sin(x!2)
Chapter 4 Solutions
( ) for 1 x
d. x for x ! 0
! 2 , i.e., for 0.5 ! x > 0
for x ! 2 (and x cannot be 2 + n! for integers n).
Calculus

140. a. y = 3(x2 !4)
x+2
= 3(x!2)(x+2)
x+2
= 3(x ! 2) (for x = !2 ) and has derivative of 3, with
y(1) = !3 ; tangent line is y + 3 = 3(x ! 1) or y = 3x ! 6 .
b. y + 3 = ! 1 3 (x ! 1)!!" y = ! 1 3 x + 1 3 ! 3!!"!!y = ! 1 3 x ! 8 3
141. a. It is hottest after ~7 months on July 1 st and coldest after ~1 month on February 1 st , by
looking at the maximum and minimum peaks on the graph.
b. The inflection points are about 4 months in, around May 1 st , and about 10 months in,
around October 1 st .
142. The derivative of each piece must be equal at x = 1; d dx ((x + 2)2 ! 3) = 2(x + 2) is 6 at
x = 1 , so d (a sin(x ! 1) + 6) = a cos(x ! 1) must be 6 at x = 1 ; a cos(0) = 6, a = 6 .
dx
143. f(x) is increasing in [!3, 2] ! [4, "] (where f ! is non-negative), f(x) is decreasing in
(!", !3] ! [2, 4] (where f ! is non-positive), f(x) is concave up in about
(!", !1.2] ! [3.1, ") (where f !(x) increases), and f(x) is concave down in about
[!1.2, 3.1] (where f !(x) decreases).
144. The 2 n indicates that the width of the interval is 2 units. The function is y = x2 . Possible
3
answers include x 2 2
! dx and ! (x + 1) 2 dx .
1
0
145. a. A: first rectangle is 9 ! (!8x + 9) = 8x by !x , and the second is 9 ! x 2 by "x .
B: !x by y " y"9 = y + y"9
8 8
C: x + x!9 by "x
8
b. Adam broke the region into two integrals:
1 3
! 8xdx + (9 " x 2 )dx
0 ! = 4x 2 1
1
( 0 ) + "
9x ! x3 3 %
3
#
$ 1 &
' = 4 + (27 ! 9) ! (9 ! 1 3 ) = 13 1 3
Becky integrated over y-values:
9
" ( y + y!9
8 ) dy = y3/2
1
3/2 + y2
9
16 ! 9 8 y = 27
312 + 81
16 ! 81
8 ! ( 2 3 + 1
16 ! 9 8 ) =
1
18 ! 5 1
16 ! !19 = 13 +
16
48 48 = 13 1 3
Cathy reflected the region across the line y = x and integrated over x-values:
9
" ( x + x!9
8 ) dx , essentially the same integral as Becky’s, = 13 1
1
3
Each method is valid and gives the same answer.
Chapter 4 Solutions
Calculus

16
146. The three possibilities are " (4 ! x )dx, " y 2 dy, and ! x 2 dx , and each equals
64
3 = 21 1 3 .
0
4
0
4
0
147. The curves intersect at 2 x = x!!!!!2 = x !!!!!x = 4 .
4
" (2 x ! x)dx = 4 3 x3/2 ! 1 2 x2 4 = 4 4
0 3 3 ! 1 2 (42 ) = 32 3 ! 8 = 8 3 = 2 2 3 un2
0
148. The functions cross at x = ! . So students must take
!
" (sin x + sin x)dx + # (! sin x ! sin x)dx or double the first integral:
0
!
2"
"
2 " (sin x + sin x)dx . (Or 4 " (sin x + sin x)dx works too). All methods yield 8 un 2 .
0
! /2
0
149. a. Using a calculator, the minimum of f is at (4, –13.5), so this is 13.5 in. This can also
be obtained by factoring the derivative:
f !(x) = x3
128 + 3x2
32 " 2 = 1 (x + 8)(x + 8)(x ! 14) is 0 at x = 4
128
b. It is ! 2
225 x2 + 20 ! x4
( 512 + x3
32 ) ! 2x ! 8 = ! x4
512 ! x3
32 ! 2x2
225
+ 2x + 28 by !x .
c. The bounds are obtained using a calculator: ! "15.326 and ! 9.753
9.753
Area ! # " x4
512 " x3
32 " 2x2 + 2x + 28 = ! x5
225
"15.326
( ) dx
! 544.60 in 2
5"512 ! x4
4"32 ! 2x3
3"225 + x2 + 28x
9.753
!15.326
1
150. a. x 2 1 #
" dx = 2 xdx
!1 " = 2 x2 1 &
0
2
$
% 0 '
( = 1, using symmetry to take double the second part
of the integral (or two integrals may be used), followed by the Fundamental Theorem.
b. 8 ! x4 4 " 1 2 ! x2 2 + C = 2x2 " x2 4
+ C , by antidifferentiating each term.
c.
5 (3x+1)(x!2)
dx
1 3x+1
5
" = " (x ! 2)dx = x2 1
2 # !2 1
5
=
25
( ) = 4
2 ! 10 ! 1 2 ! 2
d. y + C , by the second part of the Fundamental Theorem.
151. Possible answers:
5
a. f (x) = x, g(x) = cos x b. f (x) = x 3 , g(x) = 3x cos(x 2 )
c. f (x) = x + 1, g(x) = 0 d. f (x) = x + 1, g(x) = x ! 1
Chapter 4 Solutions
Calculus

152. lim
x!3 " f (x) = 4 " 3(3) = 5 , so a(3)2 + 2 = 5, 9a = 3, a = 1 3
153. y ! = "4x , so y !(1) = "4 , so the linearization is y = !4(x ! 1) + 2 = !4x + 6 .
Therefore y(1.15) ! "4(1.15) + 6 = 1.4
154. a. The left bound is 0 and the right bound is ! 2.163 (using a calculator)
2.163
" (sin x ! (x ! 1) 4 + 1)dx = ! cos x ! (x!1)5 + x
5
0
2.163
0
! 2.296 " (" 4 5 ) ! 3.096
b. The left bound is 0 and the right bound is ! 3.532 (using a calculator)
3.532
" ( x ! x(x ! 3))dx = x3/2
3/2 ! x3 3 + 3x2
3.532
2 # 8.451 ! 0 = 8.451.
0
0
155. y = x 2 ! 1
156. a. y = 2x 2
b. y = 0
c. y = x2 !3x!10
x 2 = 1 + !3x!11
!1 x 2 , so it is y = 1. The idea is that as x ! 0 or x ! " , the
+1
terms with a higher power of x in the denominator will approach zero.
157. a. lim
x!"
y = "
b. lim
x!"
y = 0
c. lim y = lim
x!" x 2 = 1
+1
These limits are the same as the limits of the respective end-behavior functions.
x!"
1+ #3x#11
158. Answers may vary but should be below 2. One method would be to estimate it as a
quadratic.
159. a. x 1 = 1
b. x 2 ! 0.7 ; x 2 is a better approximation
c. x 3 ! 0.6
d. Each successive estimate is better than the previous one.
Chapter 4 Solutions
Calculus

160. a. f !(x) = 3x 2 + 2x ; f (1) = 1 , f !(1) = 5 , tangent is y = 5(x ! 1) + 1 = 5x ! 4 .
x 2 = 0.8 , f (0.8) = 0.152 , f !(0.8) = 3.52 ,
tangent is y = 3.52(x ! 0.8) + 0.152 = 3.52x ! 2.66
x 3 = 2.664
3.52 ! 0.757 , f (x 3 ) ! 0.006 , f "(x
3 ) ! 3.232 ,
tangent is y ! 3.232(x " 0.757) + 0.006 ! 3.23x " 2.440 , x 4 ! 0.755
b. x n keeps getting smaller and closer to the actual root.
161. If f (x) = x 2 "
! N , then x n+1 = x n ! (x n )2 !N
#
$ 2xn
%
&
' = x n ! x n
2 + N
2xn = x n
2 + N
2xn = 1 2
( x n + N xn
) .
162. a. Since f (x) is a continuous function, f (!1) = !0.4597 , and f (0) = 1, by the
Intermediate Value Theorem there must be a root between x = –1 and x = 0.
b. f !(x) = 1" sin x , x 2 = "1" f ("1)
f ("1) # "0.750 , x 3
!
f ("0.750)
# "0.750 "
f !("0.750) # "0.739
c. The root is ! "0.739 (using a calculator), and the error ! 0.00003 ! 0 .
163. f !(x) = 48x 2 " 48x + 12 ; e.g. for x 1 = 0.2 , x 2 ! 0.0685 , x 3 ! 0.100 , x 4 ! 0.103 " x 5
164. a. 5 sin(6x)
b. sin(x ! 3x) = sin(!2x)
c. cos 2 (1 ! sin 2 x) ! sin 2 x(1 ! cos 2 x) =
cos 2 x ! cos 2 x sin 2 x ! sin 2 x + sin 2 x cos 2 x = cos 2 x ! sin 2 x = cos(2x)
Or, (cos 2 x ! sin 2 x)(cos 2 x + sin 2 x) = (cos 2 x ! sin 2 x) "1 = cos(2x)
d.
sin x
cos x + cos x
sin x = sin2 x+cos 2 x
cos x!sin x
=
1
1 2 sin 2x = 2 csc(2x)
165. a. Let y = sin(x + 1) . Then 2y 2 ! y ! 1 = 0 = (2y + 1)(y ! 1) , y = 1, ! 1 2
So sin(x + 1) = 1 or ! 1 2 ; x + 1 = ! 2 + 2!n or 7! 11!
+ 2!n or
6 6 + 2!n ;
7!
11!
( ) + 2!n, ( ) + 2!n, ( ) + 2!n for integers n.
x = ! 2 " 1
6 " 1
6 " 1
b. Let y = x 2 ! 2x + 1 . Then y 2 ! 3y = !2 , y 2 ! 3y + 2 = 0 = (y ! 2)(y ! 1), y = 1, 2 .
So x 2 ! 2x + 1 = 1 , x 2 ! 2x = 0 = x(x ! 2) , x = 0, 2 .
Or, x 2 ! 2x + 1 = 2 = (x ! 1) 2 , x ! 1 = ±2 , x = 3, !1 ; x = !1, 0, 2, 3
Chapter 4 Solutions
Calculus

2/(x+h)
166. lim
2 "2/x 2
h!0
h
= lim 2 # x2 "(x+h) 2
h!0 hx 2 (x+h) 2 = lim 2 #
"2xh"h 2
h!0 hx 2 (x+h) 2 =
lim 2 " #2x#h
h!0 x 2 (x+h) 2 = 2 ! "2x
x 2 !x 2 = "4x"3 = " 4
x 3
167. minimums: (!4.49, !0.217) and (4.49, !0.217)
The maximum does not exist because y(0) is undefined (is discontinuous) at x = 0 .
Inflection points: (!5.940, !0.057) , (5.940, !0.057) , (!2.082, 0.42) , (2.082, 0.42) .
168. This is when y !! is non-negative: y ! = 3x 2 + 6x " 24 , y !! = 6x + 6 ; on [!1, ") .
Chapter 4 Solutions
Calculus