MATH102 Integration: Answers to 2011 end-of-module test
1. N/A.
2. For the partial fractions, we have
Z
Then the integral
R
0
1
(x+2)(5x+5)
dy
dx
=
sin t
1cos t
dx
(x+2)(5x+4)
5
5x+4
5x + 4
x+2
converges to 16 log 25 .
1
x+2
. The indefinite integral
+ c.
MATH102 Integration: Answers to 2012 end-of-module test
1. N/A.
5x+4
x+2
1
2. For the partial fractions, we have (x+3)(x
2 +2) = x2 +2 x+3 . For the indefinite integral, we
have
2
Z
5x + 4
1
x +2
x
dx
=
log
+
2 tan1 + c.
2
2
(x + 3)(x + 2)
2
(x + 3)
2
Th
MATH102 Integration: Answers to 2013 end-of-module test
1. N/A.
2. a) The Laplace transform of x2 ex is
2
,
(s1)3
and it converges for s > 1.
b) i) f (R) diverges as R .
ii) g() log 12 = log 2 as tends to 0 from above.
3. i) False: the gradient (by implic
MATH111 Numbers and Relations
End-of-module test, November 2015
The test will last 40 minutes. Attempt all questions. Write your tutors
name on the first sheet and your own name on each sheet. The number in
brackets at the end of each question is the numb