Math 21B
1 (50 pts.)
Brian Osserman
Practice Exam 3
Evaluate the following integrals (consider them as improper integrals where
appropriate, and state clearly when you are doing so):
(a)
Z
√
x 1 − cos 2x dx.
Solution:
Z
Using the identity 1 − cos 2x = 2 sin2 x, we get
√
x 1 − cos 2x dx =
which is
Z
p
√ Z
2
x 2 sin x dx = 2 x| sin x| dx,
√ R
2 x sin x dx where sin x ≥ 0, which is on 0 ≤ x ≤ π (and
also 2π ≤ x ≤ 3π, 4π ≤ x ≤ 5π , etc.) Using integration by parts with
u = x, dv = sin x dx gives
Z
√
√
√
√ Z
2 x sin x dx = 2(−x cos x− (− cos x) dx) = − 2x cos x+ 2 sin x+C
on these intervals. On the remaining intervals, sin x ≤ 0, so | sin x| =
√ R
√
− sin x, so the integral is − 2 x sin x dx, and we get 2x cos x −
√
2 sin x + C .
Z
x2 dx
(b)
.
(x − 1)(x2 + 2x + 1)
Solution: Use partial fractions: rst, factor x2 + 2x + 1 = (x + 1)2 .
Then, solve
x2
A
B
C
=
+
+
2
(x − 1)(x + 1)
x − 1 x + 1 (x + 1)2
A(x + 1)2 + B(x − 1)(x + 1) + C(x − 1)
=
(x − 1)(x + 1)2
(A + B)x2 + (2A + C)x + (A − B − C)
=
,
(x − 1)(x + 1)2
so we want A + B = 1, 2A + C = 0, A − B − C = 0. Adding all
three equations gives 4A = 1, so we get A = 1/4, and then B = 3/4,
C = −1/2. So
Z
Z
x2 dx
3/4
1/2
1/4
=
+
−
dx
2
(x − 1)(x + 2x + 1)
x − 1 x + 1 (x + 1)2
= 1/4 ln |x − 1| + 3/4 ln |x + 1| + 1/2
(c)
Z
0
2
1
+ K.
x+1
x+1
√
dx.
4 − x2
Solution:
Note that this is an improper integral, since the integral
goes to ∞ as x approaches 2. We rst integrate
b
Z
0
x+1
√
dx =
4 − x2
Z
0
b
x
√
dx +
4 − x2
Z
b
√
0
1
dx,
4 − x2
and using the substitution u = 4 − x2 for the rst integral, we get
4−b2
Z
4
ib
√ 4−b2
−1
b
−1 x
√ du + sin
= − u 4 + sin−1 − sin−1 (0)
2 0
2
2 u
√
√
= − 4 − b2 + 4 + sin−1 (b/2)
√
= − 4 − b2 + 2 + sin−1 (b/2).
If we then take the limit as b → 2, we get
√
π
− 0 + 2 + sin−1 (1) = 2 + .
2
(d)
Z
7 dx
.
(9x2 + 1)2
Solution:
Use the trig substitution x = 13 tan θ, then 9x2 + 1 = sec2 θ
and dx = 31 sec2 θ dθ, so
Z
Z
7
sec2 θ
7 dx
=
dθ
(9x2 + 1)2
3
sec4 θ
Z
7
=
cos2 θ dθ
3
Z
7
1 + cos 2θ dθ
=
6
7
1
=
θ + sin 2θ + C
6
2
7
1
−1
−1
=
tan (3x) + sin(2 tan (3x)) + C.
6
2
2
(e)
Z
2
−∞
5 dx
.
x2 + 4
Solution:
By denition, this is
Z
lim
a→−∞
2 (20 pts.)
a
2
5 dx
=
x2 + 4
i2
5
−1 x
lim
tan
a→−∞ 2
2 a
a
5
lim tan−1 (1) − tan−1
=
2 a→−∞
2
5 π π 15π
=
− −
=
.
2 4
2
8
Consider the integral
Z
2
sin x2 dx.
0
(a) Use the trapezoidal rule with n = 4 to estimate this integral. (You do
not have to express your answer as a single fraction or decimal)
Solution:
In this case, ∆x = (2 − 0)/4 = 1/2, so the trapezoidal rule
gives
1/2
(sin 0 + 2 sin(1/4) + 2 sin 1 + 2 sin(9/4) + sin 4).
2
(b) Using the error bound formula, what is an upper bound for the error
of the estimate from (a)?
Solution:
The second derivative of sin x2 is the derivative of 2x cos x2 ,
which is 2 cos x2 − 4x2 sin x2 . The absolute value on [0, 2] is bounded
by |2 cos x2 | + |4x2 sin x2 | ≤ 2 + 16 = 18, since cos x2 and sin x2 are
always between −1 and 1. Then the (absolute value of) the error is
bounded by
3 (10 pts.)
18(b − a)3
18 · 8
3
=
= .
2
12n
12 · 16
4
Sketch the graph of the polar coordinate curve
r = cos 2θ.
3
Solution:
This is symmetric about the x-axis, since cos 2θ = cos −2θ, and
because cos 2(θ + 2π) = cos 2θ, it is enough to consider θ going from 0 to
π , and then reect the result across the x-axis. Between 0 and π/2, we see
that r goes from 1 to 0 (at θ = π/4) to −1, and then between π/2 and π we
see that r goes from −1 to 0 (at θ = 3π/4) to 1. So for y ≥ 0, we get the
top halves of two loops in opposite directions along the x-axis. For y ≤ 0
we get a single loop going in the negative direction along the y -axis. When
we reect across the x-axis, we end up with four loops, each going distance
of 1, with one each in the positive and negative x and y directions.
4 (20 pts.)
Find the area of the region inside the outer loop and outside the inner loop
of the polar coordinate curve r = 2 cos θ + 1.
Solution:
If you don't remember this curve from class, you would want to
sketch it rst. It looks like
4
The outer loop occurs when r ≥ 0, which is from θ = 0 to θ = 2π/3 and
then from θ = 4π/3 to θ = 2π ; it is more convenient to express this as θ
goes from −2π/3 to 2π/3. The inner loop occurs when r ≤ 0, which is from
θ = 2π/3 to θ = 4π/3. Because the ranges from θ do not match, and the
inner loop is completely contained inside the outer loop, it is easiest to nd
the area by nding each one separately, and substracting. For the outer
loop, observing that the integrand is an even function, we get
1
2
Z
2π/3
Z
2
2π/3
4 cos2 θ + 4 cos θ + 1 dθ
(2 cos θ + 1) dθ =
−2π/3
0
Z
=
2π/3
2(1 + cos 2θ) + 4 cos θ + 1 dθ
0
= 3θ + sin 2θ + 4 sin θ]02π/3
6π
4π
2π
=
+ sin
+ 4 sin
−0
3
3
√ 3 √
3
3
= 2π −
+4
2√
2
3 3
= 2π +
.
2
5
For the inner loop, the integrand is the same, so we get that the area is
1
2
Z
4π/3
2π/3
1
4π/3
(3θ + sin 2θ + 4 sin θ]2π/3 )
2
1 12π
8π
4π 6π
4π
2π
=
+ sin
+ 4 sin
−
− sin
− 4 sin
2
3
3
3
3
3
3
1
2π
4π
2π − 3 sin
+ 3 sin
=
2
3
3
√
3 3
= π−
.
2
(2 cos θ + 1)2 dθ =
(This matches the answer from lecture) The area inside the outer loop but
outside the inner one is then the dierence
√
√
√
3 3
3 3
2π +
−π+
= π + 3 3.
2
2
6