Here is quiz 6.
Problem 1 is Exercise 7.7.64 from the book. It asks to determine the convergence of
There are several ways of approaching this problem.
Method 1: Begin by noting that
can be evaluated by using the substitution so which transforms the integral into
that can be solved with the trig. substitution so and
Now we return to the problem:
We have found the value of the integral, so in particular, it converges.
Method 2: First,
We use comparison for both and
we have by comparison that is finite.
we have by comparison that is also finite.
Method 3: As before,
The change of variables transforms into
which of course is the same as
So all we need to do is to show that converges, and for this we can use comparison, because
Problem 2 is an easier version of Exercise 7.7.63 from the book. It asks to determine the convergence of
The problem here is that we cannot find so we are forced to use comparison. A natural thing to try would be to compare with
Since we have
The problem is that has an asymptote at 0. However, we can split the integral as
The first integral is finite, simply because it is the integral of a continuous function on some finite interval, there is nothing improper here. The second converges by comparison with the -integral with
It follows that the whole integral converges.
A slightly different approach is to use limit comparison rather than direct comparison:
It follows that converges because does; and we have to treat
as above. We cannot just compare with (which happens to diverge), because the limit comparison test requires that the functions that we compare are continuous, and is not continuous at 0.
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