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# Assignment 9 Solutions.pdf

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Western University

Mathematics

Mathematics 0110A/B

Chris Brandl

Fall

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MATH 127 Fall 2012
Assignment 9
Topics: Antiderivatives, de▯nite/inde▯nite integrals, substitution rule, Riemann sums, and
approximate integration
Due: Wednesday November 28th
Instructions:
▯ Print your name and I.D. number at the top of the ▯rst page of your solutions, and underline
your last name.
▯ Submit your solutions in the same order as that of the questions appearing.
▯ Your solutions must have legible handwriting, and must be presented in clear, concise and
logical steps that fully reveal what you are doing.
▯ If you get help or collaborate with someone, then acknowledge the names of those who helped
you. Any outright copying of assignments will be reported as an act of academic plagiarism.
1. Find the following antiderivatives.
!
Z x + x + 2
(a) tan (x) + dx
x2
2
Since we do not readily know the antiderivative of tan (x) we use the trig identity
tan (x) + 1 = sec (x). Using the identity and simplifying we get
3
2 x2 + x + 2 2 1 1 2
tan (x) + x2 = sec (x) ▯ 1 +p x + x + x2
From this we see that
!
Z 2 Z ▯ ▯
tan (x) + x + x + 2 dx = sec (x) ▯ 1 +p1 + 1+ 2 dx
x2 x x x2
p 2
= tan(x) ▯ x + 2 x + lnjxj ▯ + C
x
1 Z
1 + sin (x)
(b) 2 dx
sin (x)
Simplifying we get
1 + sin (x) 2
2 = csc (x) + sin(x)
sin (x)
From this we see that
Z 3 Z ▯ ▯
1 + sin (xdx = csc (x) + sin(x) dx
sin (x)
= ▯cot(x) ▯ cos(x) + C
Z
(c) p dx
2x ▯ x2
2
First consider the expression 2x ▯ x . If we complete the square we get
2x ▯ x = ▯(x ▯ 1) + 1
So then we have that
1 1
p = p
2x ▯ x2 1 ▯ (x ▯ 1)
Let u = x ▯ 1, we get that du = dx. This substitution gives
Z Z
dx du
p 2x ▯ x2 = p1 ▯ u2
= arcsin(u) + C
= arcsin(x ▯ 1) + C
Z e2x
(d) dx
e ▯ 1
x x
Let u = e ▯ 1 so then du = e dx. Notice that we can re-write
e dx = e e dx = (u + 1)du
We then get
Z 2x Z
e u + 1
e ▯ 1 dx = u du
Z ▯ ▯
1
= 1 + du
u
= u + lnjuj + C
= e ▯ 1 + lnje ▯ 1j + C
2 ▯ ▯
Z cos 1
(e) x dx
x2
Let u = 1 so then du = ▯ 1 dx. We get that
x x2
Z ▯1▯ Z
cos x dx = ▯cos(u)du
x2
= ▯sin(u) + C
▯ 1▯
= ▯sin + C
x
2. Evaluate the following integrals.
Z 2
(a) (x ▯ 3) dx
0
Z 2 Z 2
2 2 4 2
(x ▯ 3) dx = (x ▯ 6x + 9)dx
0 ▯0 ▯▯
x5 6x 3 ▯2
= ▯ + 9x ▯
5 3 0
5
= 2 ▯ 2(2 ) + 9(2) ▯ 0
5
32
= + 2
5
42
=
5
Z 4 3
(b) p x dx
x + 1
0 2
Let u = x + 1, we get that du = 2xdx. When x = 0 we get u = 1 and when
▯ ▯
x = 4 we have u = 17. Notice that xdx = x xdx = (u ▯ 1) 1 du.
2
3 This gives
Z 4 3 Z 17
x u ▯ 1
p 2 dx = p du
0 x + 1 Z1 ▯2pu ▯
17 u 1
= ▯ p du
1 2 2 u
! ▯17
2u 2 p ▯
= ▯ u ▯
2(3) ▯
! ▯ 1
3 ▯7
u2 p ▯
= 3 ▯ u ▯▯
1
3 p
172 1
= 3 ▯ 17 ▯ 3+ 1
3
172 p 2
= ▯ 17 +
3 3
Z ▯
4 2 p
(c) sec (x) tan(x)dx
0
Let u = tan(x), we get that du = sec (x)dx. When x = 0 we get u = 0 and when
x = ▯ we get u = 1. This gives
4
Z ▯ Z 1
4 2 p p
sec (x) tan(x)dx =

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