Math1142: Exam 3
Solutions
Vishal Saraswat
Here is the key to the third exam. This exam covered chapters 5 and 6 from the book. Be sure you understand how
to do each problem since you will see similar problems on the final exam.
1. (9 points) Find the average value of the function f (x) =
a.
b.
c.
d.
e.
1
over e ≤ x ≤ e2 .
x(ln x)2
1
2
1
e2
e2 − e
e2
1
2(e2 − e)
None of the above
Solution:
The average value of the function f (x) =
1
2
e −e
Z
e
e2
1
x(ln x)2
over e ≤ x ≤ e2 is
1
dx
x(ln x)2
Put u = ln x then du = x1 dx.
When x = e, u = ln e = 1 and when x = e2 , u = ln e2 = 2.
Therefore
Z e2
Z 2
1
1
1
1
dx = 2
du
2
2
e − e e x(ln x)
e − e 1 u2
−1 2
1
u
= 2
e − e −1 1
2
1
−1
= 2
e −e u 1
1
−1 −1
= 2
−
e −e 2
1
1 1
= 2
e −e 2
1
=
2e(e − 1)
Ans: d.
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
1
2e(e−1)
Page 1 of 9
Math1142: Exam 3
Solutions
Vishal Saraswat
2. (9 points) Find the area of the region bounded by the curves y = 8/x, y =
a. 8 ln 8 −
√
x, the line x = 8 and the x-axis.
83/2
3/2
83/2
− 8 ln 8
3/2
16
c.
+ 8 ln 2
3
16
d.
− 8 ln 2
3
e. None of the above
b.
Solution:
At the point of intersection the y values must be the same. Therefore
√
8
√x = x
or x x = 8
or x3/2 = 8
or
x = 82/3 = 4
√
Thus the curves y = 8/x and y = x meet at x = 4. We draw a rough sketch of the graph of
the two curves to determine the subdivision of the integral.
y
x=4
x=8
5
4
√
y=2 x
3
2
1
−x
O
1
−y
x
2
3
4
5
6
7
8
y = 8/x
9
√
Therefore the area of the region bounded by the curves y = 8/x, y = x, the line x = 8 and
the x-axis is
3/2 4
Z 8
Z 4
√
8
x
x dx +
dx =
+ [8 ln |x|]84
3/2 0
0
4 x
2 3/2
=
4 − 03/2 + 8 [ln 8 − ln 4]
3
2
= 8 + 8 ln(8/4)
3
16
=
+ 8 ln 2
3
Ans: c.
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
16
3
+ 8 ln 2
Page 2 of 9
Math1142: Exam 3
Solutions
Vishal Saraswat
3. (9 points) Find the particular solution of the following differential equation that satisfies the given condition:
ln x
dy
=
; y = 100 when x = 1.
dx
y
p
2(x ln x − x − 5001)
p
b. y = 2(x ln x − x + 5001)
p
c. y = 2(x ln x − x − 10001)
p
d. y = 2(x ln x − x + 10001)
a. y =
e. None of the above
Solution:
Given
dy
ln x
=
dx
y
so that on separating variables we get
y dy = ln x dx
Therefore
Z
Z
y dy =
ln x dx
and integrating both sides, we get
1 2
y = x ln x − x + C
2
on integrating by parts where C is the constant of integration. Now, solving for y we get
p
y = 2(x ln x − x + C)
From the given condition, 100 =
solution is
p
2(1 ln 1 − 1 + C) so that C = 5001. Hence the required
p
y = 2(x ln x − x + 5001).
p
Ans: b. y = 2(x ln x − x + 5001)
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
Page 3 of 9
Math1142: Exam 3
Solutions
Vishal Saraswat
4. (9 points) If the function
(
Ae−kx
f (x) =
0
if x ≥ 0
else.
is a probability density function and if f (0) = 10, find the value of k.
a. 1
b. 1/10
c. −1/10
d. −10
e. None of the above
Solution:
Since f (0) = 10, we get
10 = f (0) = A · e−k·0 = A · e0 = A · 1 = A
Since the function f (x) is a probability density function, we get
Z ∞
1=
f (x) dx
−∞
Z ∞
10e−kx dx
=
0
Z N
= 10 lim
e−kx dx
N →∞
0
N
e−kx
= 10 lim
N →∞
−k
−kN 0 −k·0 e
e
= 10 lim
−
N →∞
−k
−k
1 e−kN
= 10 lim
−
N →∞ k
k
1
= 10
−0
k
10
=
k
Therefore k = 10.
Ans: e. k = 10
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
Page 4 of 9
Math1142: Exam 3
Solutions
Vishal Saraswat
5. (9 points) Find the probability P (1 ≤ x ≤ 2) for the probability density function
(
k(3 − x) if 0 ≤ x ≤ 3
f (x) =
0
else.
a. 1/3
b. 1/9
c. 2/9
d. 2/3
e. None of the above
Solution:
The function f (x) is a probability density function. Therefore,
Z ∞
f (x) dx
1=
−∞
Z 3
=
k(3 − x) dx
0
3
x2
= k 3x −
2 0
= k (3 · 3 − 32 /2) − (0)
9
=k
2
Therefore k = 2/9. Hence
Z
2
P (1 ≤ x ≤ 2) =
f (x) dx
1
Z
=
=
=
=
=
=
=
2
2
(3 − x) dx
1 9
2
2
x2
3x −
9
2 1
2
(3 · 2 − 22 /2) − (3 · 1 − 12 /2)
9
2
[(6 − 2) − (3 − 1/2)]
9
2
[4 − 5/2]
9
2
[3/2]
9
1
3
Ans: a. 1/3
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
Page 5 of 9
Math1142: Exam 3
Solutions
Z
6. (20 points) Integrate:
Solution:
Vishal Saraswat
ln 3x ln x3
+
dx
x
x
Put u = ln 3x. Then
du =
1
1
· 3 dx = dx.
3x
x
Therefore,
Z
Z
ln 3x
u2
(ln 3x)2
dx = u du =
+C =
+C
x
2
2
Put v = ln x3 . Then
dv =
where C is the constant of integration.
1
3
· 3x2 dx = dx.
3
x
x
Therefore,
Z
Z
dv
ln x3
1 v2
(ln x3 )2
dx = v
=
+C =
+C
x
3
3 2
6
where C is the constant of integration.
Therefore,
Z
Z
Z
(ln 3x)2 (ln x3 )2
ln 3x ln x3
ln 3x
ln x3
+
dx =
dx +
dx =
+
+C
x
x
x
x
2
6
where C is the constant of integration.
Another way of doing would be to simplify the integrand by using the logarothmic rules.
ln(3x · x3 )
ln 3x4
4 ln 3x
ln 3x ln x3
+
=
=
=
x
x
x
x
x
Put u = ln 3x. Then
du =
Therefore,
Z
1
1
· 3 dx = dx.
3x
x
Z
Z
Z
4 ln 3x
ln 3x
u2
ln 3x ln x3
+
dx =
dx = 4
dx = 4 u du = 4 + C
x
x
x
x
2
2
= 2(ln 3x) + C
where C is the constant of integration.
Ans: 2(ln 3x)2 + C
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
Page 6 of 9
Math1142: Exam 3
Solutions
Vishal Saraswat
7. (20 points) Find the area of the region bounded by the curves y = x3 + 3x2 and the line y = 4x.
Solution:
At the point of intersection the y values must be the same. Therefore
x3 + 3x2
or
x3 + 3x2 − 4x
or x(x + 4)(x − 1)
or
x
=
=
=
=
4x
0
0
−4 , 0 or 1
y = x3 + 3x2
y = 4x
3
Therefore the curves y = x3 + 3x2 and y = 4x meet at
2
x = −4, x = 0 and x = 1
and hence the points of intersection are
y
4
1
−x
(−4, −16), (0, 0) and (1, 4).
−4
x
−3
−2
−1
O
1
−1
We draw a rough sketch of the graph of the two curves to
determine the subdivision of the integral.
−2
−3
Therefore the area of the region bounded by the curves
y = x3 + 3x2 and the line y = 4x is
Z 0
Z 1
3
2
[(x + 3x ) − (4x)] dx + [(4x) − (x3 + 3x2 )] dx
−4
−4
−5
0
2
0
1
x4
x
x3
x2
x4
x3
=
+ 4 −
+3 −4
−3
4
3
2 −4
2
4
3 0
(−4)4
14
3
2
3
2
=0−
+ (−4) − 2(−4) + 2 · 1 −
−1 −0
4
4
3
= 0 + 32 +
4
= 32.75
−6
−7
−8
−9
−10
Ans: 32.75
−11
−12
−13
−14
−15
x = −4
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
−16
−y
x=1
Page 7 of 9
Math1142: Exam 3
Solutions
Z
8. (20 points) Integrate:
Solution:
Vishal Saraswat
3
x5 ex dx
Let u = x3 . Then x = u1/3 and
1 1
1 2
dx = u 3 −1 du = u− 3 du
3
3
Therefore, substituting from above
Z
Z
1 2
5 x3
x e dx = (u1/3 )5 eu u− 3 du
3
(substituting from above)
Z
1
ueu du
3
Z
Z
Z d
1
u
u
(u) ·
u e du −
e du
du
=
3
dx
(integrating by parts)
Z
1
=
ueu − eu du
3
1
= {ueu − eu + C}
3
where C is the constant of integration
n
o
1 3 x3
3
=
x e − ex + C
3
(substituting back).
=
Ans:
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
1
3
3 x3
x e −e
x3
+C
Page 8 of 9
Math1142: Exam 3
Solutions
9. (20 points) Find the general solution of the differential equation:
Solution:
Given
Vishal Saraswat
dy
= 0.02xy
dx
dy
= 0.02xy
dx
so that on separating variables we get
1
dy = 0.02x dx
y
Therefore
Z
1
dy =
y
Z
0.02x dx
and on integrating both sides, we get
ln y = 0.02
x2
+C
2
where C is the constant of integration. Now, solving for y we get
y = e0.01x
2 +C
= K e0.01x
2
where K = eC is a constant.
Ans: K e0.01x
Vishal Saraswat
Math1142: Short Calculus, Spring 2010
2
Page 9 of 9