We approximate the indicated integral using the trapezoidal, midpoint and Simpson's Rule with n = 4.

• A table of values of `sqrt(1+x^4) vs. x is given.
• The second figure below shows this table and the trapezoidal approximation graph corresponding to the integral.

The trapezoidal approximation uses each 'endpoint altitude' only once and all the 'interior altitudes' twice each, always averaging the two altitudes of each trapezoid and multiplying the average by the width of the interval.

• The result is that the trapezoidal approximation is obtained by multiplying the width of the interval with half the sum of the endpoint altitudes and the sum of the interior altitudes, as indicated below.

The midpoint altitudes are shown in the third figure below, and each is used as the altitude for estimating the area over its interval.

• The resulting midpoint approximation is indicated below.

As seen in the preceding class, the errors of the left, trapezoidal and Simpson's Rule approximation tend to be invesely proportional to the first, second and fourth powers of the number of steps or subdivisions.

It follows that n2 vs. n1 subdivisions gives us

• n1 / n2 times the error for the left (or right) approximation
• (n1 / n2)^2 times the error for the trapezoidal (or midpoint) approximation
• (n1 / n2)^4 times the error for the Simpson's Rule approximation.

As an example of how we might apply these proportionalities, suppose that for some function over given interval n = 10 gives us the approximate value 3.8 when the actual integral has value 5.

• The error is therefore 1.2.
• It follows that if we increase the number of subdivisions to n = 30, then

• If our approximation was a left approximation, our error tends to change by factor n1 / n2 = 10 / 30 to 10 / 30 * 1.2 = .4.
• If our approximation was a midpoint approximation, our error tends to change by factor (n1 / n2)^2 = (10 / 30)^2 to (10 / 30)^2 * 1.2 = .13.
• If our approximation was a Simpson's Rule approximation, our error tends to change by factor (n1 / n2)^4 = (10 / 30)^4 to (10 / 30)^4 * 1.2 = .015.

If we attempt to evaluate the integral shown in the figure below, whose corresponding graph is shown in the upper right-hand corner of the figure, we find that we get an undefined term in our result.

• As shown by the graph, the function we are attempting to integrate has a vertical asymptote at x = 2, which makes it possible but not certain that the indicated area under the curve is unbounded.
• If we look at the antiderivative function ln( | x - 2 | ) shown in the second graph, we see that this function is undefined at x = 2.
• Instead of integrating from 1 to 2, we find the limiting value of the integral from 1 to x, as x approaches 2.
• We see that since the limit of ln( | x - 2 | ) is -infinity, we must say that the integral has value -infinity.

We contrast this with the integral indicated below, which also has an asymptote at x = 2 and which we therefore might expect to be represented by unbounded area.

• It turns out that the integral behaves just fine.
• This is because the antiderivative function -2 `sqrt( 2 - x ) remains finite.

If we wish to determine the gravitational potential energy change as an object escapes the gravitational field of a planet, we must perform the integral below.

• If the integral remains finite, then it is possible to escape the gravitational field of a planet. Otherwise it is not.
• Evaluating the integral 'to infinity' as a limit of integral 'to x', as x -> infinity, we obtain the result shown below.
• We conclude that escape from the gravitational field of a planet is indeed possible.

To determine whether the integral shown below converges or not, we compare it with the integral of 1 / x^2:

• We know that the indicated integral of 1 / x^2 converges.
• This fact will not be changed if we multiply the integral by 5.
• Since the function 5 / (x^2 + 7) is always less than 5 / x^2 (the denominator of the first is always > that of the second), then the integral of the first function is always less thanthat of the second.
• Since the integral of the second function converges, the integral of the first must therefore also converge.

We can use an integral to determine the volume of the pyramid shown in the figure below.

We will set up a Riemann sum that approaches the desired volume.

We think of slicing the pyramid into thin horizontal slices. We wish to find a formula for the approximate volume of slice # i.

• We start by determining the approximate base area the ith horizontal slice of the pyramid at height yi, with thickness `dyi.
• We denote the width of the slice by si.
• The approximate volume of the slice will be the area of base of slice * thickness of slice = si^2 * `dyi (approx).

In the figure below, at the bottom of the figure we see that si can be determined by constructing a side view of the pyramid, and using similar triangles.

• From the similar triangles we determine that si = 200 - 2 yi.
• It follows (see upper right-hand corner) that the approximate volume of the slice is Vi = (200 - 2 yi)^2 * `dyi.

In the figure below we write the Riemann sum that totals all the volume contributions from all n intervals, from altitude 0 to altitude 100.

• In the limit this sum approaches the indicated integral, which is easily evaluated.
• You should check to make sure that the result given is correct.