MATH 18.01 - MIDTERM 4 - SOME REVIEW PROBLEMS WITH
SOLUTIONS
18.01 Calculus, Fall 2014
Professor: Jared Speck
Problem 1. Approximate the integral
Z
2
x4 dx
0
using first Simpson’s rule with two equal intervals and then the trapezoid rule with two equal
intervals. Compare your approximations to the exact value.
Problem 2. Evaluate the integral
Z
ex sin x dx.
Problem 3. Evaluate the integral
Z
dx
.
(1 − x2 )5/2
Rx
Rx y
Problem 4. Let F (x) = 1 cosy y dy and G(x) = 1 sin
dy. Show that there is a constant C such
y2
sin x
that F (x) − G(x) = x + C. What is the value of C?
Problem 5. Evaluate the integral
Z
dx
p
.
x (ln x)2 − 4
1
2
MATH 18.01 - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS
Solutions
Problem 1. Approximate the integral
2
Z
x4 dx
0
using first Simpson’s rule with two equal intervals and then the trapezoid rule with two equal
intervals. Compare your approximations to the exact value.
Solution: We set f (x) = x4 and ∆x =
Z
2
2−0
2
= 1. Then the Simpson approximation is
∆x
[f (0) + 4f (1) + f (2)]
3
1
= (0 + 4 × 1 + 16) = 20/3.
3
x4 dx ≈
0
The trapezoid approximation is
Z 2
1
1
4
x dx ≈ ∆x f (0) + f (1) + f (2)
2
2
0
= 1 × (0 + 1 + 8) = 9.
The exact value of the integral is
Problem 2. Evaluate the integral
R2
0
x4 dx =
Z
x5
5
|x=2
x=0 = 32/5.
ex sin x dx.
Solution: We integrate by parts with u = ex , du = ex , dv = sin x dx, v = − cos x, which leads to
Z
Z
Z
x
e sin x dx = u dv = uv − v du
Z
x
= −e cos x + ex cos x dx.
Similarly, we compute that
Z
x
x
e cos x dx = e sin x −
Inserting this identity for
Z
Z
ex sin x dx.
R
ex cos x dx back into the previous formula, we deduce that
Z
x
x
e sin x dx = −e cos x + ex cos x dx
Z
x
x
= −e cos x + e sin x − ex sin x dx.
R
We then add ex sin x dx to both sides of the previous equation, divide by two, and add the
integration constant C, which leads to
Z
1
ex sin x dx = {ex sin x − ex cos x} + C.
2
MATH 18.01 - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS
3
Problem 3. Evaluate the integral
Z
dx
.
(1 − x2 )5/2
Solution: We use the inverse trig substitution x = sin u, dx = cos u du, the identity (cos u)2 =
1 − (sin u)2 , and the identity (sec u)2 = 1 + (tan u)2 , which leads to
Z
Z
cos u
du = (sec u)4 du
5
(cos u)
Z
= [1 + (tan u)2 ](sec u)2 du.
We then make the substitution v = tan u, dv = (sec u)2 du, which leads to
Z
Z
2
2
[1 + (tan u) ](sec u) du = 1 + v 2 dv
v3
+C
3
tan3 u
= tan u +
+C
3
x
1
x3
=
+
+ C.
(1 − x2 )1/2 3 (1 − x2 )3/2
=v+
In the last step above, we used a right triangle to deduce that tan u =
x
.
(1−x2 )1/2
Rx
Rx y
Problem 4. Let F (x) = 1 cosy y dy and G(x) = 1 sin
dy. Show that there is a constant C such
y2
sin x
that F (x) − G(x) = x + C. What is the value of C? Solution: We use integration by parts in
R
R
the form u = y1 , du = − y12 dy, dv = cos y dy, v = sin y, u dv = uv − v du to deduce that
Z x
Z x
cos y
sin y y=x
sin y
dy =
|y=1 +
dy
y
y
y2
1
1
|
{z
}
F (x)
sin x
=
− sin 1 +
x
Z
|1
x
sin y
dy .
y2
{z
}
G(x)
Hence,
F (x) − G(x) =
sin x
− sin 1,
x
and C = sin 1.
Problem 5. Evaluate the integral
Z
dx
p
.
x (ln x)2 − 4
4
MATH 18.01 - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS
Solution: We first make the substitution u = ln x, du = dx
, which leads to the integral
x
Z
du
.
2
(u − 4)1/2
We then make the inverse trigonometric substitution u = 2 sec θ, du = 2 sec θ tan θ dθ and use the
identity (sec θ)2 − 1 = (tan θ)2 , which leads to
Z
Z
sec θ tan θ
du
dθ
=
2
1/2
(u − 4)
tan θ
Z
= sec θ dθ
= ln |sec θ + tan θ| + C,
u r u2
= ln +
− 1 + C
2
4
r
ln x
2
(ln
x)
= ln +
− 1 + C.
2
4
√
In the next to last line above, we used a right triangle to deduce that tan θ =
u2 −4
2
=
q
u2
4
− 1.