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

Section 4.9 Antiderivatives 353
In applications of calculus it is very common to have a situation as in Example 2,
where it is required to find a function, given knowledge about its derivatives. An equation
that involves the derivatives of a function is called a differential equation. These will
be studied in some detail in Chapter 9, but for the present we can solve some elementary
differential equations. The general solution of a differential equation involves an arbitrary
con stant (or constants) as in Example 2. However, there may be some extra conditions
given that will determine the constants and therefore uniquely specify the solution.
Figure 2 shows the graphs of the function
f 9 in Example 3 and its antiderivative
f. Notice that f 9sxd . 0, so
f is always increasing. Also notice that
when f 9 has a maximum or minimum,
f appears to have an inflection point.
So the graph serves as a check on our
calculation.
40
fª
_2 3
f
FIGURE 2
_25
Example 3 Find f if f 9sxd − e x 1 20s1 1 x 2 d 21 and f s0d − 22.
SOLUtion The general antiderivative of
is
f 9sxd − e x 1 20
1 1 x 2
f sxd − e x 1 20 tan 21 x 1 C
To determine C we use the fact that f s0d − 22:
f s0d − e 0 1 20 tan 21 0 1 C − 22
Thus we have C − 22 2 1 − 23, so the particular solution is
f sxd − e x 1 20 tan 21 x 2 3
Example 4 Find f if f 0sxd − 12x 2 1 6x 2 4, f s0d − 4, and f s1d − 1.
SOLUtion The general antiderivative of f 0sxd − 12x 2 1 6x 2 4 is
f 9sxd − 12 x 3
3 1 6 x 2
2 2 4x 1 C − 4x 3 1 3x 2 2 4x 1 C
Using the antidifferentiation rules once more, we find that
f sxd − 4 x 4
4 1 3 x 3
3 2 4 x 2
2 1 Cx 1 D − x 4 1 x 3 2 2x 2 1 Cx 1 D
n
To determine C and D we use the given conditions that f s0d − 4 and f s1d − 1. Since
f s0d − 0 1 D − 4, we have D − 4. Since
f s1d − 1 1 1 2 2 1 C 1 4 − 1
we have C − 23. Therefore the required function is
f sxd − x 4 1 x 3 2 2x 2 2 3x 1 4
n
If we are given the graph of a function f, it seems reasonable that we should be able to
sketch the graph of an antiderivative F. Suppose, for instance, that we are given that
Fs0d − 1. Then we have a place to start, the point s0,1d, and the direction in which we move
our pencil is given at each stage by the derivative F9sxd − f sxd. In the next example we use
the principles of this chapter to show how to graph F even when we don’t have a formula
for f. This would be the case, for instance, when f sxd is determined by experimental data.
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