CALCULUS I HOMEWORK 11
Not Due
1. Compute the antiderivative
(a)
f (x) = 3x(2 − x)24
Solution: Let u = 2 − x. Then du = −dx and x = 2 − u. Making the
substitution,
Z
Z
24
3x(2 − x) dx =
3(2 − u)u24 (−du)
Z
Z
24
= −6 u du + 3 u25 du
u26
u25
+3
+C
25
26
6
3
= − (2 − x)25 + (2 − x)26 + C
25
26
(b) Compute the antiderivative
4x
f (x) =
1 + 4x
= −6
Solution: Let u = 1 + 4x . Then du = (ln 4)4x dx. Making the substitution
Z
Z
1 du
4x
dx
=
1 + 4x
ln 4 u
1
=
ln u + C
ln 4
1
ln(1 + 4x ) + C
=
ln 4
(c) Compute the antiderivative
f (x) =
ln x2
,
x
1
x>0
2
CALCULUS I HOMEWORK 11
Solution: Let u = ln x2 = 2 ln x. Then du = 2 dx
x . Making the substitution
Z
Z
1
ln x2
dx =
u du
x
2
1 2
=
u +C
4
1
=
(ln x2 )2 + C
4
(d) Compute the antiderivative
x+1
f (x) =
, x > −2
x+2
Solution: Let u = x + 2. Then du = dx and x = u − 2. Making the
substitution
Z
Z
(u − 2) + 1
x+1
dx =
du
x+2
u
Z 1
=
1−
du
u
= u − ln u + C
= x + 2 − ln(x + 2) + C
(e) Compute the antiderivative
2x2
1 + 2x2
f (x) =
Solution: There is a trick to deal with these types of integrals. You can
add and subtract 1 to the numerator.
Z
Z
2x2
2x2 + 1 − 1
dx
=
dx
1 + 2x2
1 + 2x2
Z 2
2x + 1
1
=
−
dx
1 + 2x2 1 + 2x2
Z 1
√
=
1−
dx
1 + ( 2x)2
√
1
= x − √ arctan( 2x) + C
2
2. Compute the derivative
(a)
Z
f (x) =
2
x2
2
e−t dt
CALCULUS I HOMEWORK 11
3
Solution: By the Chain Rule and the Fundamental Theorem of Calculus,
Z x2
d
2
4 d
4
e−t dt = e−x
x2 = 2xe−x .
dx 2
dx
(b)
Z x2
dt
√
f (x) =
2 + t4
x
Solution: Let c be any number between x and x2 . By the Chain Rule and
the Fundamental Theorem of Calculus,
Z c
Z x2
Z x2
dt
dt
d
dt
d
√
√
√
=
+
dx x
dx
2 + t4
2 + t4
2 + t4
x
c
Z x
Z x2
dt
dt
d
d
√
√
+
= −
dx c
2 + t4 dx c
2 + t4
1
1
= −√
+√
(2x)
4
2+x
2 + x8
(c)
Z e x2
f (x) =
ln(t)dt
e−x
2
Solution: Let c be any number between e−x and ex . By the Chain Rule
and the Fundamental Theorem of Calculus,
Z c
Z ex2
Z e x2
d
d
ln(t)dt =
ln(t)dt +
ln(t)dt
dx e−x
dx
e−x
c
Z e−x
Z ex2
d
d
= −
ln(t)dt +
ln(t)dt
dx c
dx c
d
d 2
2
= − ln(e−x ) e−x + ln(ex ) ex
dx
dx
−x
3 x2
= −xe + 2x e
3. Compute the integral
(a) Compute the integral
Z
π
cos x dx
−π
4
CALCULUS I HOMEWORK 11
Solution: Integrating
Z π
−π
π
cos x dx = sin x
−π
= sin π − sin(−π)
= 0
(b) Compute the integral
Z
2
√
x 2x − 1 dx
1
Solution: Let u = 2x − 1. Then du = 2dx and x = 21 (u + 1). Making the
substitution
Z u=3
Z x=2
√
√ du
1
(u + 1) u
x 2x − 1 dx =
2
u=1 2
x=1
Z u=3
1 3/2
(u + u1/2 ) du
=
4
u=1
1 3 2 5/2 2 3/2
=
u + u
4 1 5
3
1 2 5/2 2 3/2
1 2 5/2 2 3/2
=
3 + 3
−
1 + 1
4 5
3
4 5
3
1 2 1/2 1 1/2
4
=
3 3 + 33 −
10
6
15
4
7√
3−
=
5
15
(c) Compute the integral
Z π
2
sin x cos(cos x) dx
0
Solution: Let u = cos x. Then du = − sin xdx. Making the substitution
Z x= π
Z u=0
2
sin x cos(cos x) dx = −
cos udu
x=0
u=1
Z 1
=
cos u du
0
1
= sin u
0
= sin(1)
CALCULUS I HOMEWORK 11
5
(d) Compute the integral
Z
1
0
dx
√
(1 + 2 x)2
√
√
Solution: Let u = 1 + 2 x. Then du = x−1/2 dx and x = 21 (u − 1). It
√
follows that dx = xdu = 12 (u − 1)du. Making the substitution
Z x=1
Z u=3
dx
1 1
√ 2 =
(u − 1) du
22
u
(1
+
2
x)
x=0
u=1
Z
1 3 −1
(u − u−2 ) du
=
2 1
1 3
−1
=
ln u + u
2 1
1
1
1
=
ln 3 +
−
ln 1 + 1
2
3
2
ln 3 1
=
−
2
3
(e) Compute the integral
Z π
ln(1 + ex )
ex
dx
1 + ex
0
1
x
Solution: Let u = ln(1 + ex ). Then du = 1+e
x e dx, Making the substitution
Z x=π
Z u=ln(1+π)
x
x ln(1 + e )
e
dx =
u du
1 + ex
x=0
u=ln 2
ln(1+π) 2
u
= 2
ln 2
1
2
2
=
(ln(1 + π)) − (ln 2)
2
4. Find the limit
1
lim
h→0 h
Solution: Use l’Hopital’s rule.
Rh
cos tdt
lim 0
h→0
h
Z
h
cos tdt.
0
=
=
lim
d
dh
Rh
0
h→0
lim cos h
h→0
= 1.
cos tdt
1
6
CALCULUS I HOMEWORK 11
5. Let
Z
F (x) =
x
2x
sin t
dt.
t
At what value of x in the interval (0, 3π
2 ) does F (x) attain a local minimum?
Solution: Taking the derivative
Z c
Z 2x
sin t
sin t
d
0
dt +
dt
F (x) =
dx
t
t
x
c
Z x
Z 2x
sin t
sin t
d
−
dt +
dt
=
dx
t
t
c
c
sin x
sin 2x
+2
= −
2x x
1
sin 2x − sin x
=
x
Setting F 0 (x) = 0 gives the equation
sin x = sin 2x = 2 sin x cos x.
If sin x = 0, the only solution in (0, 3π
2 ) is x = π. If sin x 6= 0, we must solve
1
cos x = .
2
3π
π
The solution in (0, 2 ) is x = 3 . Hence the critical points of F (x) in (0, 3π
2 )
are x = π and x = π3 .
• On (0, π3 ), F 0 > 0 so F is increasing (can test x = π4 )
• On ( π3 , π), F 0 < 0 so F is decreasing (can test x = π2 )
π
0
• On (π, 3π
2 ), F > 0 so F is decreasing (can test x = π + 4 )
It follows that x = π is a local minimum.