lista_3__mais_integrais_imediatas

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420 ❙ ❙ ❙ ❙ CHAPTER 5 INTEGRALS
and so Equation 8 becomes
(a) If is even, then so Equation 9 gives 
(b) If is odd, then and so Equation 9 gives
Theorem 7 is illustrated by Figure 4. For the case where is positive and even, part (a)
says that the area under from to is twice the area from to because of
symmetry. Recall that an integral can be expressed as the area above the -axis
and below minus the area below the axis and above the curve. Thus, part (b) says
the integral is because the areas cancel.
EXAMPLE 10 Since satisfies , it is even and so
EXAMPLE 11 Since satisfies , it is odd 
and so
y
1
�1
 
tan x
1 � x 2 � x 4
 dx � 0
f ��x� � �f �x�f �x� � �tan x���1 � x 2 � x 4 �
 � 2[17 x 7 � x]0
2
� 2(1287 � 2) � 2847
 y
2
�2
 �x 6 � 1� dx � 2 y
2
0
 �x 6 � 1� dx
f ��x� � f �x�f �x� � x 6 � 1
0
y � f �x�
xxba f �x� dx
a0a�ay � f �x�
f
y
a
�a
 f �x� dx � �y
a
0
 f �u� du � y
a
0
 f �x� dx � 0
f ��u� � �f �u�f
y
a
�a
 f �x� dx � y
a
0
 f �u� du � y
a
0
 f �x� dx � 2 y
a
0
 f �x� dx
f ��u� � f �u�f
y
a
�a
 f �x� dx � y
a
0
 f ��u� du � y
a
0
 f �x� dx9
7–44 |||| Evaluate the indefinite integral.
7. 8.
9. 10.
11. 12.
14.
15. 16.
17. 18. y y 3s2y 4 � 1 dyy s4 � t dt
y 
1
�5t � 4�2.7
 dty 
3
�2y � 1�5
 dy
y 
x
x 2 � 1
 dxy 
dx
5 � 3x
13.
y 
x
�x 2 � 1�2
 dxy 
1 � 4x
s1 � x � 2x 2
 dx
y �2 � x�6 dxy �3x � 2�20 dx
y x
2�x 3 � 5�9 dxy 2x�x 2 � 3�4 dx
1–6 |||| Evaluate the integral by making the given substitution.
1.
2.
4.
5.
6.
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y e
sin � cos � d�, u � sin �
y 
4
�1 � 2x�3
 dx, u � 1 � 2x
y 
sin sx
sx
dx , u � sx
y x
2
sx 3 � 1 dx, u � x 3 � 13.
y x�4 � x 2 �10 dx, u � 4 � x 2
y cos 3x dx, u � 3x
|||| 5.5 Exercises
0
y
x_a a
FIGURE 4
(a) ƒ even, j   ƒ dx=2 j ƒ dx
0
a
_a
a
0
x
_a
a
y
(b) ƒ odd, j   ƒ dx=0
_a
a
SECTION 5.5 THE SUBSTITUTION RULE ❙ ❙ ❙ ❙ 421
19. 20.
22.
23. 24.
25. 26.
28.
29. 30.
31. 32.
34.
35. 36.
37. 38.
39.
40.
42.
43. 44.
; 45–48 |||| Evaluate the indefinite integral. Illustrate and check that
your answer is reasonable by graphing both the function and its
antiderivative (take ).
45. 46.
47. 48.
49–70 |||| Evaluate the definite integral, if it exists.
49. 50.
51. 52.
53. 54. y
1�2
1�6
 csc �t cot �t dty
�
0
 sec2�t�4� dt
y
s�
0
 x cos�x 2 � dxy
1
0
 x 2�1 � 2x 3 �5 dx
y
7
0
 s4 � 3x dxy
2
0
 �x � 1�25 dx
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y tan
2� sec2� d�y sin
3x cos x dx
y 
x
sx 2 � 1
 dxy 
3x � 1
�3x 2 � 2x � 1�4
 dx
C � 0
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y 
x 2
s1 � x
 dxy 
x
s
4 x � 2
 dx
y 
x
1 � x 4
 dxy 
1 � x
1 � x 2
 dx41.
y sin t sec
2�cos t� dt
�c � 0, a � �1�y x a sb � cx a�1 dx
y s
3 x 3 � 1 x 5 dxy sec
3x tan x dx
y 
sin x
1 � cos2x
 dxy cot x dx
y 
cos���x�
x 2
 dxy scot x csc
2x dx33.
y 
e x
e x � 1
 dxy 
dx
x ln x
y 
ax � b
sax 2 � 2bx � c
 dxy 
z2
s
3 1 � z 3
 dz
y e
cos t sin t dty e
x
s1 � e x dx27.
y �1 � tan ��5 sec2� d�y cos � sin6� d�
y sx sin�1 � x 3�2� dxy 
cos st
st
 dt
y 
tan�1x
1 � x 2
 dxy 
�ln x�2
x
 dx21.
y sec 2� tan 2� d�y sin � t dt 55. 56.
58.
59. 60.
61. 62.
63. 64.
66.
67.
69. 70.
; 71–72 |||| Use a graph to give a rough estimate of the area of the
region that lies under the given curve. Then find the exact area.
71. ,
72. , 
73. Evaluate by writing it as a sum of 
two integrals and interpreting one of those integrals in terms 
of an area.
74. Evaluate by making a substitution and
interpreting the resulting integral in terms of an area.
Which of the following areas are equal? Why?
76. A bacteria population starts with 400 bacteria and grows at a
rate of bacteria per hour. How many
bacteria will there be after three hours?
77. Breathing is cyclic and a full respiratory cycle from the begin-
ning of inhalation to the end of exhalation takes about 5 s. The
maximum rate of air flow into the lungs is about 0.5 L�s. This
explains, in part, why the function has
often been used to model the rate of air flow into the lungs.
Use this model to find the volume of inhaled air in the lungs at
time .t
f �t� � 12 sin�2� t�5�
r�t� � �450.268�e1.12567t
y=2x´
0 x
y
1
y=esin x sin 2x
0 x
y
π
2
1
y=eœ„x
0 x
y
1
75.
x
1
0 xs1 � x
4 dx
x
2
�2 �x � 3�s4 � x 2 dx
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
0 � x � �y � 2 sin x � sin 2x
0 � x � 1y � s2x � 1
■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■
y
a
�a
 xsx 2 � a 2 dxy
a
0
 xsx 2 � a 2 dx �a � 0�
y
a
0
 xsa 2 � x 2 dx68.y
4
0
 
dx
�x � 2�3
y
1�2
0
 
sin�1x
s1 � x 2
 dxy
e4
e
 
dx
xsln x
65.
y
4
0
 
x
s1 � 2x
 dxy
2
1
 xsx � 1 dx
y
��2
0
 cos x sin�sin x� dxy
13
0
 
dx
s
3 �1 � 2x�2
y
��2
���2
 
x 2 sin x
1 � x 6
 dxy
��3
0
 
sin �
cos2�
 d�
y
1
0
 xe�x
2
 dxy
2
1
 
e1�x
x 2
 dx57.
y
2
0
 
dx
�2x � 3�2y
��6
���6
 tan3� d�