MATH 112: SAMPLE FINAL SOLUTIONS
R5
Problem 1: Write
0
(x + 2x5 )dx as a limit of Riemann sums, using right endpoints.
Let f (x) = 2x5 + x. We want to write the integral as a sum of n rectangles with height, and then
4
take the limit as n → ∞. The width of each rectangle will be 5−1
n = n . Using right endpoints, the
height of the rectangles is f (1 + 1 · 4/n), f (1 + 2 · 4/n), . . ., f (1 + i · 4/n).
fHxL
6000
5000
4000
3000
2000
1000
x
1
lim
n→∞
n
X
(width) · (height)
2
=
i=1
=
3
4
5
n
X
4
4
f 1+i
n→∞
n
n
i=1
lim
n
X
4
lim
n→∞
n
i=1
4
2 1+i
n
5
4
+ 1+i
n
!
Problem 2:
R x arctan(t)
d
(a) Evaluate dx
e
dt
R01 arctan(x)
d
(b) Evaluate dx 0 e
dx
R 3x+1
(c) Find the derivative of the function h(x) = 2x sin(t4 )dt.
(a) Second Fundamental Theorem of Calculus:
Z x
d
f (t)dt = f (x).
dx a
Applying the theorem here we obtain simply:
Z x
d
earctan(t) dt = earctan(x) .
dx 0
(b) We claim
d
dx
To see this, note that
be zero.
R1
0
Z
1
earctan(x) dx = 0.
0
earctan(x) dx is a constant with respect to x, so the derivative will
Date: April 27, 2010.
1
2
MATH 112: SAMPLE FINAL SOLUTIONS
R 3x+1
R 2x
(c) We must split up the integral: h(x) = 0
sin(t4 )dt − 0 sin(t4 )dt. Now we may apply
the FTC. We must be careful to use the chain rule. We let u = 3x + 1, du = 3dx in the first
integral and u = 2x, du = 2dx in the second integral to obtain:
Z 3x+1
Z 2x
d
d
4
h (x) =
sin(t )dt −
sin(t4 )dt
dx 0
dx 0
Z u
Z u
d
du
du
d
=
sin(t4 )dt
sin(t4 )dt
−
du
dx
du
dx
0
0
du
du
− sin(u4 )
= sin(u4 )
dx
dx
0
=
3 sin((3x + 1)4 ) − 2 sin((2x)4 ).
Problem 3: Use the properties of integrals to show that 2 ≤
R1 √
−1
√
1 + x2 dx ≤ 2 2.
Recall that if f (x) ≤ g(x) on some interval, then the integrals of the functions satisfy the same
2
inequaliy.
On
√
√ the interval [−1, 1], we have 1 ≤ 1 + x ≤ 2. Taking square roots we obtain 1 ≤
1 + x2 ≤ 2. Integrating we obtain:
Z
1
Z
1
1dx ≤
−1
1
x ≤
−1
−1
Z 1
−1
1
Z
2≤
p
1+
x2 dx
Z
1
≤
√
2dx
−1
√
1
2x
p
1 + x2 dx ≤
p
√
1 + x2 dx ≤ 2 2,
−1
−1
and this is what we wanted to show.
y
2.0
1.5
1.0
0.5
x
-1.0
-0.5
0.0
0.5
1.0
Problem 4: Find the area bounded between the curves 4x + y 2 = 12 and x = y.
We make the substitution y = x so we may easily graph the region of interest. Next we compute the
2
intersection points of the two curves by setting them equal to each other: x = 3− x4 ⇒ x2 +4x−12 =
0 ⇒ x = −6, x = 2.
MATH 112: SAMPLE FINAL SOLUTIONS
3
y
5
x
5
-5
-5
Letting f (x) = 3 −
x2
4
and g(x) = x, we integrate the difference of the functions from −6 to 2:
Z
2
2
x2
− xdx
4
−6
x3
x2 2
= 3x −
− 12
2 −6
2
= 6 − − 2 − (−18 + 18 − 18)
3
Z
f (x) − g(x)dx =
−6
3−
2
= 22 − .
3
Problem 5: Find the volume obtained by rotating the region bounded by y = e−x , y = 0, x = −1,
x = 0 about the line x = 1.
Step 1: Draw a picture.
y
6
5
4
3
2
1
x
-3
-2
1
-1
2
3
-1
Step 2: Pick a method. Here we would like to use verticle rectangles and integrate with respect to x
from −1 to 0. This is perpendicular to the line about which we are rotating. Thus, we want
to use the method of cylindrical shells. The average radius of the cylinder will be 1 − x and
the height will be e−x − 0.
4
MATH 112: SAMPLE FINAL SOLUTIONS
Step 3: Integrate. We obtain the following integral. Note we will need to use an integration by parts
with u = x, du = dx, dv = e−x dx, v = −e−x .
Z
0
−x
2π(1 − x)e
Z
dx =
0
e−x − xe−x dx
2π
−1
−1
=
Z
0
−
0
−x 2π −e
−1
xe
−x
dx
−1
Z 0
0
0
e−x dx
2π −e−x + xe−x −
−1
−1
−1
0 0
0
−x −x −x = 2π −e + xe + e −1
−1
−1
0 = 2π xe−x =
−1
1
= 2π(−(−e ))
=
2πe.
Problem 6: Find the volume obtained by rotating the region bounded by y = 0, y = sin(x),
0 ≤ x ≤ π about the line y = 1.
Step 1: Draw a picture.
y
2.0
1.5
1.0
0.5
x
0.5
1.0
1.5
2.0
2.5
3.0
-0.5
-1.0
Step 2: Pick a method. Here we would like to use verticle rectangles and integrate with respect to x
from 0 to π. This is parallel to the line about which we are rotating. Thus, we want to use
the method of washers. The area of each washer will be A(x) = ∆xπ[(1)2 − (1 − sin(x))2 ].
Step 3: Integrate. We obtain the following integral. Note we will need the identity
cos2 (x) =
1 + cos(2x)
.
2
MATH 112: SAMPLE FINAL SOLUTIONS
π
Z
Z
π
1 − 1 + 2 sin(x) − sin2 (x)dx
A(x)dx = π
0
5
Z0 π
2 sin(x) − sin2 (x)dx
Z π
Z π
π
2 sin(x) − π
sin2 (x)dx
0
0
Z π
π
1 − cos(2x)
−2π cos(x) − π
dx
2
0
0
π
x sin(2x) π
−
−2π cos(x) − π
2
4
0
0
π
4π − π
2
2
π
4π −
.
2
= π
0
=
=
=
=
=
Problem 7: Find the length of the curve y = 16 (x2 + 4)3/2 on the interval 0 ≤ x ≤ 3.
Rp
Recall that the length of the curve is given by
1 + (dy/dx)2 dx. Here we have dy =
1/2
4) dx, so the length of the curve is given by the integral:
Z
3
r
1+
0
x
2
(x2
+
4)1/2
2
s
3
Z
dx =
1+
0
r
3
Z
1+
=
0
=
=
=
=
=
=
1
2
Z
1
2
Z
3
+
x2 2
(x + 4) dx
4
x4
+ x2 dx
4
p
x4 + 4x2 + 4dx
0
3
p
(x2 + 2)2 dx
0
Z
1 3 2
(x + 2)dx
2 0
1 x3
3
+ 2x 2 3
0
1
(9 + 6)
2
15
.
2
Problem 8: Find the area of the surface obtained by rotating the curve y =
about the x−axis.
The surface area, A, is given by:
2x
2
4 (x
√
x (for 4 ≤ x ≤ 9)
6
MATH 112: SAMPLE FINAL SOLUTIONS
y
5
4
3
2
1
x
2
4
6
8
10
-1
s
2
1 −1/2
dx
x
2
4
s
Z 9
√
1
= 2π
x 1+
dx
4x
4
Z 9s
1
= 2π
x+
dx
4
4
3/2 4π
1
9
=
x+
3
4
4
3/2 3/2 !
1
1
4π
9+
− 4+
.
=
3
4
4
Z
A =
2π
9
√
Problem 9:
Z
x 1+
2
|x2 − 4x|dx.
−2
This is an absolute value integral. Let f (x) = x2 − 4x. We are interested to know where f (x) is
positive and negative in the interval [−2, 2]. Note that f (x) has roots x = 0 and x = 4. We check
the function at a value in (−2, 0) and (0, 2):
f (−1) = (−1)2 − 4(−1) = 5
f (1) = (1)2 − 4(1) = −3.
Thus f (x) > 0 on the interval (−2, 0) and f (x) < 0 on the interval (0, 2). After splitting up the
integral with the appropriate signs, the integration is basic:
Z 2
Z 0
Z 2
|x2 − 4x|dx =
(x2 − 4x)dx −
(x2 − 4x)dx
−2
−2
0
3
3
x
x
0
2
=
− 2x2 −
− 2x2 3
3
−2
0
−8
8
= −
−8 −
−8
3
3
8
8
=
+8− +8
3
3
= 16.
MATH 112: SAMPLE FINAL SOLUTIONS
Problem 10:
Z
sec(θ) tan(θ)
dθ
1 + sec(θ)
We use a u−substitution, with u = sec(θ) + 1 and du = sec(θ) tan(θ)dθ.
Z
sec(θ) tan(θ)
dθ
1 + sec(θ)
Z
=
=
1
du
u
ln |u| + C
ln |1 + sec(θ)| + C.
=
Problem 11:
√
Z
3
arctan(1/x)dx
1
Here we use integration by parts being careful to use the chain rule:
u = arctan(1/x),
1 −1
du =
dx
1 + x12 x2
−1
= 2
dx.
x +1
dv = dx
v=x
So we obtain the following integral, and must use a u−substitution with u = x2 + 1, du = 2xdx:
√
Z
1
Problem 12:
3
√3 Z
arctan(1/x)dx = x arctan(1/x) −
1
1
√
3
−x
dx
+1
x2
√3 1 Z 1
= x arctan(1/x) +
du
2
u
1
√
√
√
1
3
2
= [ 3 arctan(1/ 3) − arctan(1)] +
ln |x + 1| 2
1
√ π π
1
= [ 3 − ] + (ln |3 + 1| − ln |1 + 1|)
6
4
2
√
3π π 1
=
− + ln(2)
6
4
2
Z
1
dx.
(1 − x2 )3/2
7
8
MATH 112: SAMPLE FINAL SOLUTIONS
Here we use a trigonometric substitution, letting x = sin θ, dx = cos θdθ, and the identity 1−sin2 θ =
cos2 θ.
Z
Z
cos θ
1
dθ
dx
=
2
3/2
(1 − x )
(1 − sin2 θ)3/2
Z
cos θ
=
dθ
(cos2 θ)3/2
Z
cos θ
=
dθ
(cos3 θ)
Z
1
=
dθ
(cos2 θ)
Z
=
sec2 θdθ
=
=
Problem 13:
Z
√
4x2
tan θ + C
x
√
+ C.
1 − x2
1
dx.
− 4x − 3
For this integral we will need to combine many techniques including completing the square, u−substitution,
and trigonometric substitution. We begin by completing the square, noting that (4x2 − 4x − 3)1/2 =
2(x2 − x − 3/4)1/2 = 2((x − 1/2)2 − 1)1/2 . Then we use a u−substitution letting u = x − 1/2,
du = dx. Finally, we will use a trig-sub letting u = sec θ, du = sec θ tan θdθ, and using the identity
sec2 θ − 1 = tan2 θ.
Z
Z
1
1
√
dx =
dx
1 2
1/2
2
2((x
−
4x − 4x − 3
2 ) − 1)
Z
1
du
=
2(u2 − 1)1/2
Z
sec θ tan θ
=
dθ
2(sec2 θ − 1)1/2
Z
sec θ tan θ
=
dθ
2(tan2 θ)1/2
Z
1
=
sec θdθ
2
1
=
ln | sec θ + tan θ| + C
2
1
=
ln |(x + 1/2) + (4x2 − 4x − 3)1/2 | + C.
2