Math 2215 - Calculus 1
Homework #22 Solutions
Assigned - 2009.10.13
Due - 2009.10.19
Textbook problems:
Section 4.1 - 1, 4, 6, 7, 10, 13, 16, 20, 22, 23, 25, 28, 33, 34, 36, 58, 59, 61
Fun Problems:
1. Verify that f (x) = tan2 (x) and g(x) = sec2 (x) have the same derivative. What can you conclude about the
relation between f (x) and g(x)? Verify this conclusion directly.
d
d
tan2 (x) = 2 tan(x) tan(x) = 2 tan(x) sec2 (x)
dx
dx
d
d
sec2 (x) = 2 sec(x) sec(x) = 2 sec(x) sec(x) tan(x) = 2 tan(x) sec2 (x)
dx
dx
So we have
d
d
tan2 (x) =
sec2 (x) = 2 tan(x) sec2 (x),
dx
dx
which implies that f (x) = g(x) + c, i.e. tan2 (x) = sec2 (x) + c. To verify this directly, remember from trig that
tan2 (θ) + 1 = sec2 (θ). So in the way we have it written, c = −1 and thus tan2 (x) = sec2 (x) − 1.
2. Find constants c1 and c2 such that F (x) = c1 x sin(x) + c2 cos(x) is the antiderivative of f (x) = x cos(x).
First we compute F 0 (x):
F 0 (x) = c1 sin(x) + c1 x cos(x) − c2 sin(x) = (c1 − c2 ) sin(x) + c1 x cos(x).
Setting F 0 (x) equal to f (x):
(c1 − c2 ) sin(x) + c1 x cos(x) = x cos(x)
note that we must have c1 = 1 and c1 − c2 = 0, i.e. c1 = c2 . This if we set c1 = c2 = 1 we should have the correct
function: F (x) = x sin(x) + cos(x), with F 0 (x) = f (x). It is easy to check that this is indeed the case.
3. Suppose F 0 (x) = f (x) and G0 (x) = g(x). Is it true that F (x)G(x) is an antiderivative of f (x)g(x)? Prove the
statement or give a counterexample to disprove the statement.
Essentially we are asking if the following statement is true:
Z
F 0 (x)G0 (x)dx = F (x)G(x)
As a simple example, let f (x) = g(x) = 1. This F (x) = G(x) = x and the above equation would read:
Z
1 · 1 dx = x · x
but we know that
Z
1 dx = x,
so the statement is not true!
1
2
4. Show that if F 0 (x) = f (x), then 12 F (2x) is an antiderivative of f (2x). For k 6= 0, can you find a formula for
the general antiderivative of f (kx) given F 0 (x) = f (x)?
If 12 F (2x) is an antiderivative of f (2x), i.e.
Z
f (2x)dx =
then
1
F (2x),
2
d 1
F (2x) = f (2x).
dx 2
Can we verify this? Let’s try:
1 d
d 1
F (2x) =
F (2x)
dx 2
2 dx
1
d
= F 0 (2x) ·
2x
2
dx
1
= F 0 (2x) · 2
2
= F 0 (2x)
= f (2x).
We therefore have shown that
1
2 F (2x)
is an antiderivative of f (2x).
In general it should be easy to modify the above string of equalities to show that
f (kx):
1 d
d 1
F (kx) =
F (kx)
dx k
k dx
1
d
= F 0 (kx) ·
kx
k
dx
1
= F 0 (kx) · k
k
= F 0 (kx)
= f (kx).
1
k F (kx)
is an antiderivative of