MATH 101 V2A
January 28th – Practice problems
Hints and Solutions
1. Integration by Parts:
Z e
(a) Evaluate
log(t)dt.
1
Solution: Let u = log(t) and dv = dt. Then du = 1t dt, v = t and using the method of parts gives
us
Z e
e Z e
e 1dt = e − t = e − (e − 1) = 1.
log(t)dt = t log(t) −
1
1
Z
(b) Evaluate
e
e2
t2
1
1
log(log(t))
dt.
t
Solution: I’ve just realized that this problem isn’t really doable. I hope you haven’t wasted too
much time on it – if you have, you might feel better in knowing that I have also wasted a lot of
time on it!!
Z
(c) Let f (t) be continuously differentiable. Prove that
b
f (t)f 0 (t)dt =
a
1
f (b)2 − f (a)2 .
2
Solution: Let u = f (t) and dv = f 0 (t)dt. Then du = f 0 (t)dt, v = f (t) and the method of parts gives us
that
Z b
b Z b
f (t)f 0 (t)dt = f (t) · f (t) −
f (t)f 0 (t)dt.
a
a
So
a
b
Z
f (t)f 0 (t)dt = f (b)f (b) − f (a)f (a),
2
a
and therefore
Z
b
f (t)f 0 (t)dt =
a
1
f (b)2 − f (a)2 .
2
You could also do this problem using the method of substitution.
2. Partial Fractions:
Z 3
6
(a) Evaluate
dt.
2
2 t −1
Hint: Factor the denominator into t2 − 1 = (t − 1)(t + 1) and use partial fractions to get that
t2
6
−3
3
=
+
−1
t+1 t−1
Use this to get that
Z
3
2
Z
−3
(b) Evaluate
−4
t4
6
dt = −3 log(2) + 3 log(3) − 3 log(4) = 3 log
2
t −1
3
.
2
6t2
dt.
− 5t2 + 4
Hint: Factor the denominator into t4 − 5t2 + 4 = (t + 2)(t − 2)(t + 1)(t − 1) and use partial fractions
to get that
6t2
2
−2
−1
1
=
+
+
+
.
4
t − 5t2 + 4
t−2 t+2 t−1 t+1
Use this to get that
3 Z −3
5
6t2
dt
=
3
log(5)
−
log(4)
+
3
log(2)
−
log(3)
−
2
log(6)
=
log
.
4
2
2 · 33
−4 t − 5t + 4
Z
1
(c) Evaluate
−1
2t3 + 11t2 + 28t + 33
dt.
t2 − t − 6
Hint: Use long division to divide the polynomials and get that 2t3 + 11t2 + 28t + 33 = (t2 − t −
6)(2t + 13) + 53t + 111. Use this to rewrite the integral as
Z
1
−1
2t3 + 11t2 + 28t + 33
dt =
t2 − t − 6
Z
1
53t + 111
dt
t2 − t − 6
Z 1
1
53t + 111
2
= t + 13t +
dt
2−t−6
t
−1
−1
Z 1
53t + 111
= 26 +
dt.
2
−1 t − t − 6
2t + 13 +
−1
Now use partial fractions on the remaining integral.
Answer: 26 + 54 log(2) − log(3) − 54 log(4) = 26 − 54 log(2) − log(3).
2