Exam 1 Solutions
Math 112 - Spring 2015
Feb 13, 2015
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Question
Points
1(a)
12
1(b)
12
1(c)
12
2(a)
12
2(b)
4
3(a)
8
3(b)
8
3(c)
8
4
12
Free
12
Total
100
Score
12
1.
(36 points) Evaluate the integral
a. (12 pts)
Z
e
ln x dx
1
Hint: You can simplify your answer to get an integer (but that’s not necessary for full credit).
We use integration by parts, with u = ln x =⇒ du = x1 dx and dv = dx =⇒ v = x.
This gives
Z
Z
Z
1
ln x dx = x ln x − x dx = x ln x −
dx = x ln x − x + C
x
Now,
Z
e
ln x dx = (e ln(e) − e) − (1 ln(1) − 1) = (e − e) − (0 − 1) = 1
1
b. (12 pts)
Z
sin2 x cos2 x dx
Since the exponents of sin and cos are both even, we half to use the half angle formulas sin2 x =
1−cos(2x)
and cos2 x = 1+cos(2x)
to reduce the exponents.
2
2
Z
Z
Z
(1 − cos(2x))(1 + cos(2x))
1
2
2
dx =
(1 − cos2 (2x)) dx
sin x cos x dx =
4
4
From here, we have to use a half angle formula a second time:
cos2 (2x) =
1 + cos(4x)
2
This gives us:
Z
Z Z
1
1
1 + cos(4x)
1
2
(1 − cos (2x)) dx =
dx =
(1 − cos(4x)) dx
1−
4
4
2
8
1
=
8
sin(4x)
x sin(4x)
x−
+C =
−
+C
4
8
32
c. (12 pts)
Z
(x2
x−3
dx
+ 1)(2x − 1)
We must first find A, B, C so that
Ax + B
C
(Ax + B)(2x − 1) + C(x2 + 1)
x−3
=
+
=
(x2 + 1)(2x − 1)
x2 + 1
2x − 1
(x2 + 1)(2x − 1)
Expanding, we have that
(Ax+B)(2x−1)+C(x2 +1) = 2Ax2 −Ax+2Bx−B+Cx2 +C = (2A+C)x2 +(−A+2B)x+(−B+C)
So we want to find A, B, C so that
2A + C = 0
−A + 2B = 1
−B + C = −3
Solving this system of linear equations gives us A = 1, B = 1, C = −2.
Now we integrate. First, we split up the constant and linear term:
Z
Z
Z
x
1
x+1
dx
=
dx
+
dx
x2 + 1
x2 + 1
x2 + 1
For the linear term, we use u-sub with u = x2 + 1 =⇒ du = 2x dx.
Z
Z
1
1
1
1
x
dx
=
du = ln |u| = ln |x2 + 1|
2
x +1
2
u
2
2
The constant term just gives an inverse trig function:
Z
1
dx = tan−1 (x)
2
x +1
Now, lastly the u-sub u = 2x − 1 =⇒ du = 2 dx gives
Z
Z
−2
1
dx = −
du = − ln |u| = − ln |2x − 1|
2x − 1
u
Putting everything together, we get that
Z
x−3
1
dx = ln |x2 + 1| + tan−1 (x) − ln |2x − 1| + C
(x2 + 1)(2x − 1)
2
2.
(16 points) Consider the integral
Z
3
3x2 dx
−1
a. (12 pts) Use the Trapezoid Rule with n = 2 to approximate it.
We need to split the interval [−1, 3] into two: [−1, 1] and [1, 3]. For each of these two intervals,
we will get a trapezoid. The formula for the area of a trapezoid is A = b(h12+h2 ) , and in our case
b = 2.
2 × (3(−1)2 + 3(1)2 ) 2 × (3(1)2 + 3(3)2 )
+
= 3 + 3 + 3 + 27 = 36
2
2
b. (4 pts)
Is this an overestimate or an underestimate? Explain why.
Solution 1:
Z
3
−1
3x2 dx = x3 |3−1 = 33 − (−1)3 = 27 + 1 = 28
So it’s an overestimate .
Solution 2: f (x) = 3x2 has f (2) = 6, so is concave up. This means that the trapezoid will
be above the curve, and so will be an overestimate.
Solutions 3: Draw a careful picture, and notice that the trapezoid is above the curve.
3.
(24 points) Let y = 13 x3/2 − x1/2 , 1 ≤ x ≤ 9
2
dy
a. (8 pts)
Find 1 + dx
Hint: Writing this expression as a square will help you later.
dy
1
1
= x1/2 − x−1/2
dx
2
2
2
1 1/2 1 −1/2 2
dy
1
1 1 −1
=1+
1+
x − x
=1+
x− + x
dx
2
2
4
2 4
1
1 1/2 1 −1/2 2
1 1 −1
= x+ + x =
x + x
4
2 4
2
2
Notice that adding one just changes the sign of the middle term, which means that we can still
write the resulting expression as a square - just using a plus sign instead of a minus sign.
b. (8 pts)
Use your answer in part (a) to find the exact length of the curve.
Z
L=
9
s
1+
1
dy
dx
2
Z
dx =
1
9
1 x1/2 + 1 x−1/2 dx
2
2
Since the expression inside the absolute value will always be positive in the interval [1, 9], we
can get rid of it:
9
Z 9
1 3/2
1 1/2 1 −1/2
1/2 x + x
dx =
x +x
L=
2
2
3
1
1
=
1
1
32
1
× 93/2 + 91/2 − × 13/2 − 11/2 = 9 + 3 − − 1 =
3
3
3
3
c. (8 pts)
Use your answer in part (a) to find the exact area of the surface obtained by
rotating the curve about the y-axis.
9
s
Note that this equals
1712π
15 ,
2
9
1 1/2 1 −1/2
A=
2πx 1 +
dx
dx = 2π
x
x + x
2
2
1
1
Z 9
1 3/2 1 1/2
1 5/2 1 3/2 9
= 2π
dx = 2π
x + x
x + x
2
2
5
3
1
1
5
1
3
1
1
1
1 1
5/2
3/2
5/2
3/2
= 2π
×9 + ×9 − ×1 − ×1
= 2π
+9− −
5
3
5
3
5
5 3
Z
dy
dx
Z
but that’s more simplification than I’d want to do by hand...
4.
(12 points) True or False? If you think it’s true, show your work. If you think it’s false,
explain why and give the correct solution.
Z 3
1
dx = ln 2
0 x−1
1
is not defined at x = 1, and therefore it is an improper integral and
The function f (x) = x−1
we can’t use the fundamental theorem of calculus to compute it directly - we must split it up
into two parts and take limits.
Z
lim
t→1+
Z
t
t
3
1
dx = lim (ln |x − 1|)|3t = ln |2| − lim ln |t − 1| = ∞
x−1
t→1+
t→1+
1
dx = lim (ln |x − 1|)|t0 = lim ln |t − 1| − ln | − 1| = −∞
t→1−
t→1−
0 x−1
R3 1
Since these improper integrals diverge, the improper integral 0 x−1
dx diverges too, and does
lim
t→1−
not equal ln 2. False
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