Math 105/206 - Quiz 3, Feb 27 2015
IMPORTANT: Write your name AND student number somewhere on this sheet.
No calculators, books or notes. Please show your work to get full marks. (10 marks total)
Problem 1
Compute the following indefinite integral using the method of partial fractions:
Z
dx
2
x − 3x + 2
Solution
The denominator factors as x2 − 3x + 2 = (x − 2)(x − 1), and by the usual method you find
x2
1
1
1
=
−
.
− 3x + 2
x−2 x−1
So
Z dx
1
1
=
−
dx
x2 − 3x + 2
x−2 x−1
= log |x − 2| − log |x − 1| + C.
Z
Problem 2
Compute the following definite integral using a trigonometric substitution:
Z 2 √
x2 4 − x2 dx
0
If you need it, you can use the formula
Z
cos2 x sin2 x dx =
x sin(4x)
−
+ C.
8
32
Solution
By setting x = 2 sin u you get dx = 2 cos u du and the limits of integration become u = 0 and u = π2 , so
Z
2
x
2
√
Z
4−
x2
dx = 16
0
π/2
cos2 u sin2 u du
0
= 16
π/2
u sin(4u) −
=π
8
32
0
Problem 3
Compute the following improper integral:
Z
0
∞
ex
dx
(1 + ex )2
Hint: use a substitution to compute the indefinite integral
Solution
By using the substitution u = ex + 1 we get du = ex dx and
Z
Z
ex
dx = u−2 du = −u−1 + C = −(ex + 1)−1 + C.
(1 + ex )2
So
Z
0
∞
Z b
ex
ex
dx
=
lim
dx
b→+∞ 0 (1 + ex )2
(1 + ex )2
= lim (−(ex + 1)−1 |b0 )
b→+∞
1
1
1
= lim − b
+
=
b→+∞
e +1 2
2
as eb + 1 tends to +∞ as b → +∞, so
1
eb +1
tends to 0.