School of Mathematics and Statistics
Carleton University
Math. 69.104
ASSIGNMENT 2
SOLUTIONS
Due: November 16, 1999 (or Nov. 15, 1999 Mr. Dubé),
Solution
a) The intercepts are found as follows: y =0 when x=-1. When x=0 we have y=1.
or when
. The graph is concave down in the interval
d) The points of inflection occur when there is a change in concavity. In this case they are found to be at
and at
.
e) There is only one critical point (a local maximum) at x=1 (since 

Now the integral breaks up into 3 pieces as follows:
Solution Use the Substitution Rule: Let
so that
. The integral now looks like,
or
Solution Use the Substitution Rule first: Let
so that
. The integral now looks like,
Bring this to a more recognizable form where the Table Method may be used..., that is, let
Then,
and
Solution The denominator factors as
where the quadratic is irreducible. Thus, since the degree of the numerator is less than the degree of the denominator, we get
where A = 1 by the cover-up method. To get B, C we simply multiply throughout by the denominator and find
since A=1. Comparing coefficients we see that B=1 and C=-1. It follows that
Integrating we get