MATH 2260 Worksheet
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Worksheet Week 8 Model Solution
Section 8.7.
• This worksheet is for improvement of your mathematical writing skill. Writing
using correct mathematical expression and steps is really important part of doing
math. Please print out this worksheet and try to solve problems, following given
steps.
• You don’t need to submit this worksheet. It is not homework.
• The steps in the solution represent important information you have to use in the
solution. It must be mentioned on your own solution.
1. Evaluate the integral
Z
0
xex dx.
−∞
(a) Find points in the interval of integral such that given function is not defined
at there.
xex is not defined at x = −∞
(b) By using the definition of improper integral, set up a limit of an integral.
Z
0
Z
x
xe dx = lim
a→−∞ a
−∞
0
xex dx
(c) Evaluate the integral we’ve found at step (b).
Z
0
xex dx
a
Z
x
f (x) = x,
g 0 (x) = ex
f 0 (x) = 1,
g(x) = ex
x
xe dx = xe −
Z
a
0
Z
ex dx = xex − ex + C
xex dx = [xex − ex ]0a = (0−e0 )−(aea −ea ) = −1−aea +ea = (1−a)ea −1
1
MATH 2260 Worksheet
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(d) Evaluate the limit we’ve found at step (b).
0
Z
a→−∞ a
= lim
a→−∞
1−a
−1
a→−∞ e−a
xex dx = lim (1 − a)ea − 1 = lim
lim
a→−∞
−1
− 1 (I used L’Hôpital’s rule here to evaluate the limit)
−e−a
=
−1
− 1 = 0 − 1 = −1
−∞
2. Evaluate the integral
1
Z
0
1
√ dx.
x
1
√ is not defined at x = 0
x
1
Z
1
√ dx = lim
x
a→0+
0
Z
a
1
1
√ dx =
x
1
Z
Z
a
1
1
√ dx
x
h 1 i1 1
1
1
1
x− 2 dx = 2x 2 = 2 · 1 2 − 2 · a 2 = 2 − 2a 2
a
a
Z
lim
a→0+
a
1
1
1
√ dx = lim 2 − 2a 2 = 2
x
a→0+
2
MATH 2260 Worksheet
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3. Evaluate the integral
Z
2
∞
x2
2
dx.
−1
2
is not defined at x = ∞
x2 − 1
Z ∞
Z a
2
2
dx = lim
dx
2
2
a→∞
x −1
2
2 x −1
x2
=
2
2
A
B
=
=
+
−1
(x − 1)(x + 1)
x−1 x+1
A(x + 1)
B(x − 1)
(A + B)x + A − B
+
=
(x − 1)(x + 1) (x − 1)(x + 1)
(x − 1)(x + 1)
⇒ A + B = 0, A − B = 2 ⇒ 2A = (A + B) + (A − B) = 0 + 2 = 2
⇒ A = 1, B = −1 ⇒
Z
Z
lim
a→∞ 2
a
a
2
dx =
2
x −1
Z
1
−1
1
1
2
=
+
=
−
x2 − 1
x−1 x+1
x−1 x+1
a
1
1
−
dx = [ln |x − 1| − ln |x + 1|]a2
x+1
2
2 x−1
a − 1
+ ln 3
= (ln |a − 1| − ln |a + 1|) − (ln |1| − ln |3|) = ln a + 1
a − 1
2
a − 1 dx = lim ln +ln 3 = ln lim
+ln 3 = ln 1+ln 3 = ln 3
a→∞
a→∞ a + 1 x2 − 1
a + 1
3
MATH 2260 Worksheet
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4. Evaluate the integral
Z
1
(− ln x) dx.
0
− ln x is not defined at x = 0
Z
1
Z
1
(− ln x) dx = lim
a→0+
0
Z
1
Z
a
a
f (x) = − ln x,
1
f 0 (x) = − ,
x
Z
(− ln x) dx = − ln x · x −
Z
a
1
1
1 · (− ln x) dx
(− ln x) dx =
Z
(− ln x) dx
a
g 0 (x) = 1
g(x) = x
1
− · x dx = − ln x · x +
x
Z
1 dx = − ln x · x + x + C
(− ln x) dx = [− ln x · x + x]1a = (− ln 1 · 1 + 1) − (− ln a · a + a) = 1 + a ln a − a
Z
a→0+
1
(− ln x) dx = lim 1 + a ln a − a = lim 1 +
lim
a→0+
a
= lim 1 +
a→0+
1
a
− a12
a→0+
− a = lim 1 − a − a = 1
a→0+
4
ln a
1
a
−a