To evaluate a definite integral using the trapezoidal approximation, we use the following formula:

Using the formula, we get

To solve the integral using the trapezoidal approximation, we must use the following formula:

Now, using the above formula, we can approximate the integral:

To solve the integral, it is easiest to first rewrite it so that the larger value is the upper bound. This is done by switching the bounds and also making the integral negative:

Now, we can use the trapezoidal approximation, which states that for a definite integral:

Using the above formula for our integral, we get

The function has a removable discontinuity at .

Since this function is undefined at  is it not continuous across any interval containing .

Notice that the correct answer is an open interval that goes up to, but does not include .

This function (shown below) is defined for every value along the interval with the given conditions (in fact, it is defined for all real numbers), and is therefore continuous.  However, there is a cusp point at (0, 0), and  the function is therefore non-differentiable at that point.  

The derivative of the funtion using the power rule

is ,

so it is not continuous when  or is negative.

This occurs when , so the interval of continuity will be when , which is the interval .

For rational expressions, points of discontinuity often occur when the denomenator equals zero (triggering a division by zero).

Here we note that the denomenator will equal zero at positive and negative 4; however, we first must factor ensure that neither of these are factorable.

When factoring the rational expression the following occurs:

We now see that the point x=4 was a removable discontinuity and in fact the only time this function is discontinous is at x=-4.

Recall: For a function to be continous at a point, the point must exist, the limit must exist, and the limit must equal the point.

The limit as any asymptotic function (a function with an asymptote) approaches infinity is always its horizontal asymptote.

To find the horizontal asymptote of a rational expression one must first look for the following conditions:

1. If the highest power of x is in the numerator then the horizontal asymptote does not exist.
2. If the highest power of x is in the denomenator then the horizontal asymptote is at 0.
3. If the highest power is in both, then divide the leading coefficents.

In the case of this problem the highest power is in both so we divide the leading coefficents

Therefore, our limit is 2.

To find relative maximums, we need to find where our first derivative changes sign. To do this, find your first derivative and then find where it is equal to zero.

Begin with:

    at 

This means we have extrema at x=0 and x=-8/3

Because we are only concerned about the interval from -5 to 0, we only need to test points on that interval. Try points between our two extrema, as well as one between -8/3 and -5.

So our first derivative goes from negative to positive at -8/3, thus this is the x coordinate of our relative maximum on this interval.

Though there are relative maxima at  and  (found by setting the first derivative of the function equal to zero and solving for x.)  the maximum value along the whole interval is actually at the upper endpoint, when .  When looking for extrema along an interval, looking for zeros of the first derivative does not account for endpoint extrema.