Here is the second midterm.

**Question 1:** *Find the volume of the solid obtained by rotating about the -axis the region between the -axis and the curve for .*

Here is a graph of in the specified region. To compute the volume, we use the shell method, and see that it is given by

We evaluate this integral using integration by parts.

We take: and so so

**Question 2:** *Solve the initial value problem , *.

We use the technique of separation of variables, and rewrite the given differential equation as:

so

To solve this integral, we begin by completing the square in the denominator of the given fraction:

This indicates that to solve the integral we want to use the trigonometric substitution

or

This gives:

We conclude by finding the value of , for which we use that This gives us

or

Putting this together, we have

**Question 3:** *Find .*

To solve this integral, we use the technique of partial fractions decomposition, and begin by factoring

This means we need to find constants and such that

As seen in class, there are several ways of doing this. For example, we can begin by adding the fractions on the right hand side, to obtain

This means that

for all values of . When this gives

When , we have

When , we have

Finally, when we have

or

Putting this together, we have

and

[Actually, although this was not addressed in lecture, we can use different values of in each of the intervals and .]

**Question 4:** *We have that . Suppose we do not know the value of . We can use the trapezoidal rule to approximate it. Find a value of such that the theoretical error for the trapezoidal rule using subdivisions is strictly smaller than . Find the approximation given by the trapezoidal rule using this .*

The error using the trapezoidal rule with subintervals is bounded above by

where and where

To compute a bound on we first find

An easy way of bounding this expression is to note that is increasing for with value at and value for so the maximum of its absolute value is Similarly, is decreasing, so it maximum occurs at , where it is 1.

Hence is bounded above by and

If we want it suffices that

or

so

The approximation is given by

which is indeed an accurate approximation to within

[Here is a graph of for A more careful computation of would have revealed that which gives (Note that but looks at the maximum of not just .)

The approximation gives the value . A less careful computation, replacing with (so we do not have to worry about both positive and negative numbers) gives , with Actually, works as well but the theoretical error does not predict this.]

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