# 1 SOLUTIONS NAME: STUDENT ID NUMBER:

```1
MATHEMATICS 105 (Section 921) - Quiz #2
SOLUTIONS
NAME:
STUDENT ID NUMBER:
SIGNATURE:
INSTRUCTIONS: Notes, books and calculators are not permitted. Answers may
receive no credit if accompanying work is not provided. When indicated, simplify
answers as much as possible, and be sure to clearly indicate the final answer for
each question. Please use the back of the pages if you need extra space.
4
1. Approximate the definite integral
Z 2
xe−2x dx
0
with the right-hand Riemann sum using n = 4 subintervals. You do not need to simplify your
Here we have f (x) = xe−2x and ∆x = b−a
n =
3
1, x3 = 2 and x4 = 2. The right-hand sum is
2−0
4
= 21 . So we have the points x1 = 12 , x2 =
S2 ( f ) = f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + f (x4 )∆x
= ∆x [ f (x1 ) + f (x2 ) + f (x3 ) + f (x4 )]
1 1 −1
3 −3
−2
−4
=
&middot;
e +1&middot;e + e +2&middot;e
2 2
2
1
1
3
1
=
+ 2 + 3 + 4.
4e 2e
4e
e
2
6
2. Compute the following integrals (3 marks each)
(a)
Z
4x − 1
dx
2 + x − x2
2 + x − x2 = (2 − x)(1 + x), so look for a partial fraction decomposition of the form
A
B
4x − 1
=
+
(2 − x)(1 + x) 2 − x 1 + x
multiply both sides by (2 − x)(1 + x),
4x − 1 = A(1 + x) + B(2 − x)
Plugging in x = −1 gives −5 = 3B so B = − 53 . Plugging in x = 2 gives 7 = 3A so A = 73 .
Hence,
Z
−5
1
1
dx +
dx
2−x
3
1+x
5
7
= − ln|2 − x| − ln|1 + x| + c.
3
3
4x − 1
7
dx
=
2 + x − x2
3
Z
Z
Note the extra negative in front of the first term comes from making the substitution u = 2 − x
(so du = −dx) to integrate the first term.
(b)
Z
x2 e−2x dx
Use IBP with u = x2 so du = 2xdx and let dv = e−2x dx, so that v = − 12 e−2x . By the IBP
formula:
Z
Z
1 2 −2x
2 −2x
x e dx = − x e + xe−2x dx
2
Use IBP again, with u = x so du = dx and let dv = e−2x dx, so that v = − 12 e−2x again. By the
IBP formula:
Z
Z
1 2 −2x
1 −2x
1 −2x
2 −2x
x e dx = − x e + − xe +
e dx
2
2
2
1
1
1
= − x2 e−2x − xe−2x − e−2x + c.
2
2
4
```