School of Mathematics and Statistics
Carleton University
Math. 69.107
SOLUTIONS TO OLD TEST 3
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PART I: Multiple Choice Questions
(Choose and CIRCLE only ONE answer)
(a)
(b)
(c)
(a)
(b)
(c)
(d) .
(a)
(c)
(d)
(a)
(b)
(d)
Use a trig. identity and a substitution
The value of the improper integral is given by I = 1
, (b) FALSE
PART II: Show all work here.
No additional pages will be accepted
a) .
Solution: Let . Then . When x=0, u=0 and when , u=0. and so
Alternately,
Thus,
b) Evaluate
Use the Table Method:
and the final answer can be written as,
[4 marks] (a)
Solution: Write
Integrate by Parts twice or use the "MYCAR" trick! You'll get
where the entry in the Box is given by . See Section 8.3.4 for details. This I is an antiderivative, and so
[4 marks] (b)
Solution: The degree of the numerator exceeds that of the denominator and so we must divide these expressions. Thus
Next, we use partial fractions on the third integral thus:
where we let , , in the second integral (and didn't have to use partial fractions).