2002 Test 3 solution

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10 Apr 2012
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MAT 137Y 2002-2003, Solutions to Term Test 3
1. Evaluate the following expressions.
(6%) (i) d
dx ln

x
1
.
f
x

ln
x
1
implies f
x

1
x
1
1
2
x.
(6%) (ii) d
dx x
1
x.
Knowing that ab
ebln a,d
dxe
ln x
x
e
ln x
x

x
x
ln x
2
x
x

x
1
x
1
1
2lnx
x3
2

(6%) (iii) f
e
, where f
x

tan
1

ln x
1
e
t2dt
. Simplify your answer.
Using FTC and the chain rule,
f
x

1
1
ln x
1
e
t2dt
2
e

ln x
2
1
x
thus f
e

1
e
1
e
1
e
2
2. Evaluate the following integrals.
(8%) (i)
e
x
xdx.
By letting u
x,
e
x
xdx
2e
x
C.
(8%) (ii)
e2xcosx dx.
Let Ibe the integral. Integrate by parts twice. Let u
e2xand dv
cosx dx. Then du
2e2xdx
and v
sinx. Hence
I
e2xsinx
2
e2xsinx dx
Now let u
e2xand dv
sinx dx, giving du
2e2xdx and v
cosx.
I
e2xsinx
2
!
e2xcosx
2
e2xcosx dx
"#
e2xsinx
2e2xcosx
4
e2xcosx dx
e2xsinx
2e2xcosx
4I
Solving for I,
I
1
5e2xsinx
2
5e2xcosx
C
1
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