James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

404 Chapter 5 Integrals
that when a formula for a general indefinite integral is given, it is valid only on an
interval. Thus we write
y 1 x dx − 2 1 2 x 1 C
with the understanding that it is valid on the interval s0, `d or on the interval s2`, 0d.
This is true despite the fact that the general antiderivative of the function f sxd − 1yx 2 ,
x ± 0, is
Fsxd −
2 1 x 1 C 1 if x , 0
2 1 x 1 C 2 if x . 0
_1.5 1.5
FIGURE 1
4
_4
The indefinite integral in Example 1
is graphed in Figure 1 for several
values of C. Here the value of C is
the y-intercept.
Example 1 Find the general indefinite integral
y s10x 4 2 2 sec 2 xd dx
SOLUTION Using our convention and Table 1, we have
y s10x 4 2 2 sec 2 xd dx − 10 y x 4 dx 2 2 y sec 2 x dx
You should check this answer by differentiating it.
Example 2 Evaluate y cos
sin 2 d.
− 10 x 5
5 2 2 tan x 1 C
− 2x 5 2 2 tan x 1 C
SOLUTION This indefinite integral isn’t immediately apparent in Table 1, so we use
trigonometric identities to rewrite the function before integrating:
y cos
sin 2 d − y S 1
sin DS cos
sin D d
− y csc cot d − 2csc 1 C
n
n
Example 3 Evaluate y 3
sx 3 2 6xd dx.
SOLUTION Using FTC2 and Table 1, we have
0
y 3
sx 3 2 6xd dx − x 4
0
4 2 6 x 2 3
2G0
− ( 1 4 ? 34 2 3 ? 3 2 ) 2 ( 1 4 ? 04 2 3 ? 0 2 )
− 81
4
Compare this calculation with Example 5.2.2(b).
2 27 2 0 1 0 − 26.75
n
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