2003 Test 3 solution

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10 Apr 2012
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MAT 137Y, 2003-2004 Test 3 Solutions
1. Compute the following integrals.
(10%) (i) Z52+xdx.
Z52+xdx =25Z5xdx =25 5x
ln5 +C.
(10%) (ii) Zx3(lnx)2dx.
Let Ibe the integral and integrate by parts. We let u= (lnx)2,dv =x3dx. Then du =2lnx
xdx
and v=1
4x4. Hence,
I=1
4x4(lnx)2Z1
2x3lnxdx
=1
4x4(lnx)21
21
4x4lnx1
4Zx3dx(u=lnx,dv =x3dx,du =1
xdx,v=1
4x4)
=1
4x4(lnx)21
8x4lnx+1
32x4+C.
(10%) (iii) Zdx
x2x2+16.
Let x=4tanu. Then dx =4sec2udu and this gives
Zdx
x2x2+16 =Z4sec2udu
16tan2u·4secu=1
16 Zsecu
tan2udu =1
16 Z1
cosu·cos2u
sin2udu
=1
16 Zcscucotudu =1
16 cscu=x2+16
16x+C.
(10%) (iv) Z4x
(x1)2(x+1)dx.
We separate the integral using partial fractions. For some constants A,B, and C:
A
x1+B
(x1)2+C
x+1=4x
(x1)2(x+1)=A(x1)(x+1) + B(x+1) +C(x1)2=4x.
If x=1, this gives 2B=4, or B=2. If x=1 we get 4C=4, or C=1. If x=0, we get
A+B+C=0, or A=1. Hence,
Z4x
(x1)2(x+1)dx =Z1
x1+2
(x1)21
x+1dx =ln|x1|2(x1)1ln|x+1|+C.
2. Compute the following limits.
(8%) (i) lim
x1+(lnx)tanπx
2.
The limit is of the form 0·(). Rewriting the limit,
lim
x1+(lnx)tanπx
2=lim
x1+
lnx
cot πx
2
H
=lim
x1+
1
x
csc2πx
2·π
2
=2
π,
since csc π
2=1.
1
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