MATH 105 101
Assignment 4 Solutions
Due date: October 30, 2014
MATH 105 101 Assignment 4 Solutions
All work must be shown for full marks.
1. (14 marks) Evaluate the following indefinite integrals:
Z √
a) (7 marks) et 9 − e2t dt
Z
2
√
b) (7 marks)
dx
x − 3 x + 10
Solution:
a) We first do a direct substitution with u = et , so du = et dt, and get:
Z √
Z √
t
e 9 − e2t dt =
9 − u2 du
Use
p substitution with u = 3 sin(v), and du = 3 cos v dv. Thus,
√ an inverse
2
9 − u = 9 − (3 sin v)2 = 3 cos v, and we get
Z √
Z
2
9 − u du = 9 cos2 v dv
Z 1 + cos(2v)
= 9
dv
(Use half angle formula)
2
Z
Z
9 cos(2v)
9
dv +
dv
=
2
2
9v 9 sin(2v)
=
+
+C
2
4
9v 9 sin v cos v
=
+
+C
(Use double angle formula)
2
2
√
9 arcsin u3
u 9 − u2
=
+
+C
2 2
t
√
9 arcsin e3
et 9 − e2t
=
+
+C
2
2
t
√
Z √
9 arcsin e3
t
e
9 − e2t
Thus, et 9 − e2t dt =
+
+ C.
2
2
b) We first do a substitution with u =
√
dx
x + 10. So, du = √
, that is,
2 x + 10
2u du = dx and x = u2 − 10. Thus,
Z
Z
2
4u
√
dx =
du
2
u − 10 − 3u
x − 3 x + 10
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MATH 105 101
Assignment 4 Solutions
Due date: October 30, 2014
Since u2 − 10 − 3u = (u − 5)(u + 2), we will use partial fractions. Suppose
that:
u2
4u
A
B
(A + B)u + (2A − 5B)
=
+
=
− 10 − 3u
u−5 u+2
u2 − 10 − 3u
Thus, we obtain A + B = 4 and 2A − 5B = 0. From the second equation, we
5B
get A =
, which we can use to substitute in the first equation to obtain
2
5B
8
20
+ B = 4. Solving for B, we get B = , and thus, A = . Therefore,
2
7
7
Z
Z
Z
20
8
4u
du =
du +
du
2
u − 10 − 3u
7(u − 5)
7(u + 2)
20 ln |u − 5| 8 ln |u + 2|
=
+
+C
7√
7
√
20 ln | x + 10 − 5| 8 ln | x + 10 + 2|
=
+
+C
7
7
√
√
Z
2
20 ln | x + 10 − 5| 8 ln | x + 10 + 2|
√
+
+ C.
Thus,
dx =
7
7
x − 3 x + 10
2. (11 marks) Determine if the following improper integrals converge or diverge:
Z ∞
2
xe−x dx
a) (5 marks)
−∞
3
Z
1
dx
2x4
b) (3 marks)
−2
3
Z
c) (3 marks)
√
0
1
dx
3−x
Solution:
a) From the definition of integrals over unbounded intervals, we have that:
Z ∞
Z 0
Z ∞
2
−x2
−x2
xe
dx =
xe
dx +
xe−x dx
( if both integrals converge)
−∞
∞
−∞
Z
xe
Z 00
xe
−∞
−x2
0
dx = lim
b→∞
−x2
b
Z
xe
dx
( if the limit exists)
0
Z
dx = lim
a→−∞
−x2
0
2
xe−x dx
a
Page 2 of 5
( if the limit exists)
MATH 105 101
Assignment 4 Solutions
Z
First, to find
Due date: October 30, 2014
2
xe−x dx, we use a direct substitution with u = −x2 so
du = −2x dx. Thus,
Z
Z
2
eu
−e−x
eu
−x2
+C
xe
dx = − du = − + C =
2
2
2
We now evaluate the limits:
Z ∞
Z b
2
−e−x b
2
−x2
xe
dx = lim
xe−x dx = lim
|0
b→∞ 0
b→∞
2
0
2
−e−b
1
1
= lim
+ =
b→∞
2
2
2
Z 0
Z 0
2
−e−x 0
−x2
−x2
|a
xe
dx = lim
xe
dx = lim
a→−∞ a
a→−∞
2
−∞
2
1
1 e−a
=−
= lim − +
a→−∞
2
2
2
Since
Z ∞ both integrals converges, we can conclude that the improper integral
2
xe−x dx also converges to their sum, which is 0.
−∞
1
b) Over the interval [−2, 3], the function f (x) = 4 is continuous everywhere
2x
except x = 0 where it is not defined. So, we use the definition of improper
integral:
Z 3
Z 0
Z 3
1
1
1
dx =
dx +
dx
( if both integrals converge)
4
4
4
−2 2x
−2 2x
0 2x
Z 0
Z b
1
1
dx = lim−
dx
( if the limit exists)
4
4
b→0
−2 2x
−2 2x
Z 3
Z 3
1
1
dx = lim+
dx
( if the limit exists)
4
4
a→0
0 2x
a 2x
Z
1
−1
Since
dx
=
+ C, we can evaluate the limits:
2x4
6x3
Z b
Z 0
1
1
−1
dx = lim−
dx = lim− 3 |b−2
4
4
b→0
b→0 6x
−2 2x
−2 2x
−1
1
= lim− 3 −
=∞
b→0 6b
48
Z 3
Z 3
1
1
−1
dx = lim+
dx = lim+ 3 |3a
4
4
a→0
a→0 6x
0 2x
a 2x
−1
−1
= lim+
− 3 = ∞.
a→0 162
6a
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MATH 105 101
Assignment 4 Solutions
Due date: October 30, 2014
Since both
Z 3integrals do not converge, we can also conclude that the improper
1
integral
dx also diverges.
4
−2 2x
1
is continuous everywhere
3−x
except x = 3 where it is not defined. So, we need to use the definition of
improper integral:
Z 3
Z b
1
1
√
√
dx = lim−
dx
( if the limit exists)
b→3
3−x
3−x
0
0
c) Over the interval [0, 3], the function f (x) = √
Since √
√
1
dx = −2 3 − x + C, we can evaluate the limit:
3−x
Z 3
Z b
√
1
1
√
√
dx = lim−
dx = lim− −2 3 − x |b0
b→3
b→3
3−x
3−x
0
0
√
√
√
= lim− −2 3 − b + 2 3 = 2 3
b→3
Z
Thus, the improper integral
0
3
√
√
1
dx converges to the value of 2 3.
3−x
Total: 25 marks.
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