Download Antiderivatives

Antiderivatives
Can we find a function F (x) whose derivative is x2 ? That is, if
F 0 (x) = x2 ,
can we find F (x)? Since F 0 (x) is a power, we may expect F (x) to be a power function.
Furthermore, computing the derivative of a power function causes the power to decrease
by one, so a first guess may be
F (x) = x3 .
0
A quick check shows this is wrong, x3 = 3x2 , not x2 . A pesky 3 popped out, so we try
modifying our guess:
1
F (x) = x3 .
3
This time it checks out:
0
1 3
= x2 .
x
3
This illustrates the process of antidifferentiation: We are given a function f (x) and we
wish to find a function F (x) so that
F 0 (x) = f (x).
f (x) is the derivative of F (x). We also say F (x) is an antiderivative of f (x). Antidifferentiation is the process of reversing differentiation.
One thing to notice in that last definition: we said F (x) is AN antiderivative of f (x),
not THE antiderivative of f (x). This suggests f (x) may have more than one antiderivative.
Indeed, going back to our example, F (x) = 13 x3 − 19 is another antiderivative of x2 since
0
1 3
x − 19 = x2 .
3
As is F (x) = 13 x3 + 197364, or for that matter F (x) = 31 x3 + C, where C is any fixed
real number. Once we find one antiderivative of f (x), we can generate infinitely many
antiderivatives simply by adding a constant to our first antiderivative.
Fact 1. If F (x) is an antiderivative of f (x), then for any real number C, F (x) + C is also
an antiderivative of f (x).
Returning once again to our example, must every antiderivative of x2 be of the form
suppose G(x) is some antiderivative of x2 , so
1 3
3 x + C for some real number C? Let’s
G0 (x) = x2 . Also, ( 31 x3 )0 = x2 . Therefore,
1 3 0
G (x) =
x
3
for all real numbers x. The “+ C” Theorem guarantees that
1
G(x) = x3 + C
3
for some constant C, so indeed we know every possible antiderivative of x2 .
This holds rather generally, provided we restrict attention to finding antiderivatives of
functions whose domains intervals and the functions are continuous.
0
1
2
Fact 2. If f (x) is continuous on an interval I and F (x) and G(x) are two antiderivatives
of f (x), then there is a constant C so that G(x) = F (x) + C.
This fact ceases to hold if we insist on finding antiderivatives on domains other than intervals. At the end of this note we’ll look at an example hinting at the necessary modifications
for non intervals.
Most of the above discussion is rather conditional: IF we can find an antiderivative of
f (x), THEN we can find lots of antiderivatives of f (x). We still need to figure out how to
find that first antiderivative. There is no systematic procedure for finding antiderivatives
in general, but we can generate formulas for lots of antiderivatives simply by reversing
formulas for derivatives.
Example 1. Find an antiderivative of cos x.
Solution: Do we know a function whose derivative is cos x? Yes,
(sin x)0 = cos x,
so sin x is an antiderivative of cos x.
1
2x .
Example 2.
(a) Find an antiderivative of 1+x
2 − 3e
(b) Find the most general antiderivative of the function in part a.
1
2x satisfying F (0) = 2?
(c) Is there an antiderivative, F (x), of 1+x
2 − 3e
Solution: Part (a): Let’s look at each term in the difference separately. After a moment’s
reflection we recall
1
.
(arctan x)0 =
1 + x2
For the second term, we recall
(e2x )0 = 2e2x .
We want an antiderivative of 3e2x , not 2e2x . Multiplying through by an appropriate constant
we find
3 2x 0
e
= 3e2x .
2
1
2x . Let’s check:
I now claim F (x) = arctan x − 32 e2x is an antiderivative of 1+x
2 − 3e
3 2x 0
3 2x 0
1
3
0
arctan x − e
= (arctan x) −
e
=
− e2x .
2
2
2
1+x
2
1
2x is defined and continuous for all real numbers, so Fact 2 applies,
Part (b): 1+x
2 − 3e
showing that any antiderivative must be of the form
3
arctan x − e2x + C
2
for some constant C.
Part (c): Suppose F (x) is an antiderivative satisfying F (0) = 2. From [b.] we know
F (x) = arctan x − 32 e2x + C for some C. So
3
3
2 = F (0) = arctan 0 − e2·0 + C = 0 − + C.
2
2
3
Thus,
3
2 = − + C,
2
so C = 27 , so
3
7
F (x) = arctan x − e2x + .
2
2
Example 3. Find an antiderivative of (2x − 1)(8x3 − 7).
Solution: An antiderivative of the first factor is x2 − x and an antiderivative of the
second factor is 2x4 − 7x. Check:
(x2 − x)0 = 2x − 1
(2x4 − 7x)0 = 8x3 − 7.
Great! So an antiderivative of (2x − 1)(8x3 − 7) is (x2 − x)(2x4 − 7x), right? WRONG!
Taking the derivative of (x2 − x)(2x4 − 7x) does not give (2x − 1)(8x3 − 7). Remember the
product rule! The derivative does not behave nicely with respect to products, therefore we
should not expect antiderivatives to be well behaved with respect to products either. We
can get an antiderivative of (2x − 1)(8x3 − 7) by first expanding:
(2x − 1)(8x3 − 7) = 16x4 − 8x3 − 14x + 7.
An antiderivative is then
16 5 8 4 14 2
x − x − x + 7x.
5
4
2
(Check that its derivative is actually 16x4 − 8x3 − 14x + 7)
2
Example 4. Find an antiderivative of 2xex .
Solution: After much trial and error we may finally stumble across the antiderivative
2
ex .
Why? Well,
2
ex
2
0
2
2
= (x2 )0 ex = 2xex .
2
This verifies that ex is an antiderivative of 2xex , but it gives no insight to how we might
2
come up with ex . In the second semester you will learn a general method that applies to
problems like this.
Example 5. Is there an antiderivative, F (x), of 2x−3 satisfying F (1) = 1 and F (−1) = 2?
1
Solution: One antiderivative of 2x−3 is F (x) = 2 −3+1
x−3+1 = −x−2 . What is the
most general antiderivative? Doesn’t Fact 2 imply F (x) = −x−2 + C is the most general
antiderivative? NO! Fact 2 does not apply here, since 2x−3 is defined and continuous on
(−∞, 0) ∪ (0, ∞). We may apply fact 2 on each of these intervals separately, but the most
general antiderivative is
(
−x−2 + C1 ,
x < 0;
F (x) =
−2
−x + C2 ,
x > 0.
4
Thus, we may have a different constant on each of the intervals. Now,
1 = F (1) = −1−2 + C2
so C2 = 2. Also
so C1 = 3. Therefore
2 = F (−1) = −(−1)−2 + C1
(
−x−2 + 3,
F (x) =
−x−2 + 2,
x < 0;
x > 0.
is an antiderivative of 2x−3 satisfying F (1) = 1 and F (−1) = 2.
Table of Antiderivatives
Function
c
xn
x−1
ekx
ax
sin x
cos x
sec2 x
sec x tan x
√ 1
1−x2
1
1+x2
cosh x
sinh x
Antiderivative
cx, c a constant
1
n+1
x
provided n 6= −1
n+1
ln |x|
1 kx
e
k
1 x
a
ln a
− cos x
sin x
tan x
sec x
arcsin x
arctan x
sinh x
cosh x
Each antiderivative formula in the table is obtained by taking a familiar derivative formula and “reading it backwards.” You can easily check the validity of each of these formulas
by computing the derivative of the left column; the result is the function in the right hand
column.