2004 Test 3 solution

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10 Apr 2012
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MAT 137Y 2004-2005, Solutions to Term Test 3
1. Evaluate the following integrals.
(10%) (i) Zsec2x
tanx+1dx.
Let u=tanx+1. Then du =sec2x dx. Therefore,
Zsec2x
tanx+1dx =Zdu
u=ln|u|=ln|tanx+1|+C.
(10%) (ii) Zp9x2dx.
Let x=3sinθ. Then dx =3cosθdθ. This gives
Z3cosθ·3cosθdθ=Z9cos2θdθ=9Z1+cos2θ
2dθ=9
2θ+9
4sin2θ.
Substituting back, we see that sin2θ=2sinθcosθ=2·x
3·9x2
3, so
Zp9x2dx =9
2sin1x
3+1
2xp9x2+C.
(10%) (iii) Z2x2x+4
x(x2+4)dx.
We decompose the rational function using partial fractions:
2x2x+4
x(x2+4)=A
x+Bx+C
x2+4=A(x2+4) + Bx2+Cx =2x2x+4=
A+B=2
C=1
4A=4,
so A=1, B=1, and C=1. Thus,
Z2x2x+4
x(x2+4)dx =Z1
x+x1
x2+4dx =ln|x|+1
2ln(x2+4)1
2tan1x
2+C.
2.
(10%) (i) Find the volume generated by rotating the region bounded by the curves y=ex,y=ex,x=1
about the y-axis.
Since the region is being rotated about the y-axis, we use shells. Since ex>exfor x>0, the
volume is
V=Z1
02πx(exex)dx
Integrating by parts, we let u=xand dv =exexdx. Then du =dx and v=ex+ex. Hence
V=Z1
02πx(exex)dx =2πx(ex+ex)
1
0Z1
0ex+exdx
=2πe+e1hexexi1
0
=2πe+e1e+e1=4πe1.
1
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