Math 126 - Calculus II
Lab Section 63988
Quiz 4
This is a group quiz. Write the names of your group members at the top of your quiz.
Only turn in one quiz per group. There are three questions. Point values are listed with
each question. Show all of your work for full credit. Only non graphing calculators are
allowed; no cell phones.
1. (1.5 points) Identify and perform the appropriate trig substitution. Do not solve the
integral! Stop when you have the integral in terms of your substitution variable.
Z
Z
Z
dt
dx
x2
(b)
(c)
√
dx
(a)
2
2
25t + 2
(4x − 1)3/2
9 − x2
Solution: To determine which trig sub to use, see how the equations
cos2 x + sin2 x = 1 and 1 + tan2 x = sec2 x
relate to the integral.
√
(a) The appearance of 9 − x2 means we want x = 9 sin θ = 3 sin θ as our
substitution. Then dx = 3 cos θ dθ and the integral becomes
Z
Z
9 sin2 θ
p
3 cos θ dθ = 9 sin2 θ dθ.
2
9 − 9 sin θ
(b) 25t2 + 2 ⇒ we want 5t =
and the integral becomes
Z
√
2 tan θ. Then t =
√
2 sec2 θ dθ
=
5(2 tan2 θ + 2)
√
2
5
√
tan θ and dt =
2
5
sec2 θ dθ,
Z √
2
dθ.
10
(c) 4x2 − 1 ⇒ we want 2x = sec θ. Then x = sec2 θ , dx = sec θ2tan θ dθ, and the
integral becomes
Z
Z
sec θ
sec θ tan θ
=
dθ.
2
3/2
2(sec θ − 1)
2 tan2 θ
2. Solve the following indefinite integrals:
Z
dx
(a) (1.5 points)
2
(x − 4)(x − 3)
Z
dt
(b) (1.5 points)
2
9t + 4
Z
(c) (2 points)
Z
(d) (2 points)
√
x2
dx
− 4x + 8
x3
dx
x3 − x2 + 4x − 4
Solution:
(a) Factor and use the method of partial fractions first:
1
A
B
C
=
+
+
(x + 2)(x − 2)(x − 3)
x+2 x−2 x−3
1 = A(x − 2)(x − 3) + B(x + 2)(x − 3) + C(x + 2)(x − 2)
This equation allows for a shortcut. If we set x = 2, then we find B = −1/4.
Setting x = 3 gives C = 1/5 and setting x = −2 gives A = 1/20. Now we
integrate:
Z
1
1
1
1
1
1
−
+
dx =
ln |x+2|− ln |x−2|+ ln |x−3|+C.
20(x + 2) 4(x − 2) 5(x − 3)
20
4
5
(b) Since we have 9t2 + 4, sub 3t = 2 tan θ. Then t = 32 tan θ, dt =
9t2 + 4 = 4 sec2 θ, and the integral becomes
Z
Z
1
2 sec2 θ dθ
1
=
dθ = θ + C.
2
3 · 4 sec θ
24
24
We still need to get θ in terms of t. Solve t =
1
arctan( 3t2 ) + C.
arctan( 3t2 ). So our answer is 24
2
3
2
3
sec2 θ dθ,
tan θ for θ to get θ =
(c) We can’t factor x2 − 4x + 8, so start by completing the square to get
(x − 2)2 + 4. Then
p it looks like we want to substitute x − 2 = 2 tan θ. Then
dx = 2 sec2 θ dθ, (x − 2)2 + 4 = 2 sec θ, and the integral becomes
Z
Z
2 sec2 θ dθ
= sec θ dθ = ln | sec θ + tan θ| + C.
2 sec θ
Now get the answer in terms of x. Solve x − 2 √
= 2 tan θ for tan θ and
p
(x−2)2 +4
+ x−2
| + C as
(x − 2)2 + 4 = 2 sec θ for sec θ. Then we have ln |
2
2
the answer.
(d) Since the degree of the denominator is NOT greater than the degree of
the numerator, before we can apply partial fractions, we need to do some
polynomial long division:
1
3
2
3
x − x + 4x − 4
x
− x3 + x2 − 4x + 4
x2 − 4x + 4
x3
x2 − 4x + 4
=
1
+
x3 − x2 + 4x − 4
x3 − x2 + 4x − 4
Now we can apply the method of partial fractions to the second term, once
we factor the denominator by grouping to get (x − 1)(x2 + 4) :
⇒
x2 − 4x + 4
A
Bx + C
=
+ 2
2
(x − 1)(x + 4)
x−1
x +4
2
2
x − 4x + 4 = A(x + 4) + (Bx + C)(x − 2)
x2 − 4x + 4 = (A + B)x2 + (−2B + C)x + 4A − 2C
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⇒ 1 = A + B, −4 = −2B + C, 4 = 4A − 2C
Solve the system of equations using whatever method you prefer to get
A = 1/5, B = 4/5, and C = −16/5. Going back to our integral, now we
have
Z
4x − 16
1
+
dx
1+
5(x − 1) 5(x2 + 4)
We know how to integrate the first two terms. We need to split the last term
up over subtraction:
Z
1
4x
16
1+
+
−
dx
5(x − 1) 5(x2 + 4) 5(x2 + 4)
We can integrate the third term using the regular substitution u = x2 +4, and
we can integrate the fourth term using a tangent trig sub (or by recognizing
the antiderivative of arctan). We end up with
x+
1
4 1
16 1
x
ln |x − 1| + · ln |x2 + 4| −
· arctan( ) + C.
5
5 2
5 2
2
3. (1.5 points) Set up, do not solve, an integral to calculate the amount of work (in
joules) required to pump all of the water out of a full hemispherical tank of radius
10 m, where water exits through a spout of height 2 m on the top of the tank. Use
that the density of water is 1000 kg/m3 and acceleration due to gravity is 9.8 m/s2 .
Solution: We need to fill in the components of the formula
Z
W = mass × acceleration × distance.
Since we are dealing with volumes of water, we will use that mass = density ×
volume. Sketch the situation:
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Be careful with the picture! We are looking for the volume and distance in terms
of y. We get the volume by breaking things up into tiny vertical pieces and adding
them up. Each tiny piece has volume = area of a cross sectional circle × ∆y. The
radii of the circles change depending on y, but we canp
get the radius in terms of
2
2
2
y with the equation x + (y − 10) = 10 ⇒ radius = 100 − (10 − y)2 . We are
shifting the circle up by 10 so that its bottom sits on the line y = 0. This will
simplify setting up the integral.
Having determined our volume, we find the distance from the spout to the tiny
vertical piece of water being lifted - our picture illustrates it should be 12 − y.
Then the integral is
Z 10
1000 · π(100 − (y − 10)2 ) · 9.8(12 − y) dy.
W =
0
Alternatively, if you do not shift the circle and work with the equation x2 + y 2 =
102 , the distance to lift the water out of the spout will be given by 2 − y; that
is, the distance from the top of the spout to the depth of the water (visualize the
top of the tank being at the line y = 0). Our limits of integration will be from
−10 to 0. Thus,
Z 0
W =
1000 · π(100 − y 2 ) · 9.8(2 − y) dy.
−10
These two integrals give the same amount of work.
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