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Webwork Problem 3

cbrowni1

I am having troubles with question 3 on the HW and was wondering if anyone could help me out with understanding this stuff. My version of question 3 asks me to estimate:

\int_1^5 x^5 \, dx

using both midpoint and trapezoidal sums with n=6 terms. Here’s a screen shot of the exact question:

chowell1

The question wants you to find the trapezoidal and midpoint sums of that integral. To do so, you need all of the x_i values. To find the necessary formulas on the errors page on the marksmath website.

mark

First off, note that I’ve updated the code at the bottom of our Riemann sums page. There are now buttons making it easy to switch between all 4 types of sums that we’ve learned about. That should make it pretty to get credit on the WebWork.

Of course, we want to understand what’s going on for our forum problem and for the exam. So here’s an illustrative example:

Estimate

\int_{-1}^1 e^{-x^2} dx

using a trapezoidal sum with n=12 terms.

The general formula for a trapezoidal sum is

\int_a^b f(x) \, dx \approx \sum_{i=1}^n f\left(\frac{x_i+x_{i-1}}{2}\right) \Delta x.

Note that a, b, and f are always known - typically given in the statement of the problem. The number of terms n is either known or can be computed using an error estimated. In the problem at hand, it’s given as part of the problem.

We’ve just got to figure out what \Delta x and x_i are, which is not too hard:

\Delta x = (b-a)/n = (1-(-1))/12 = \frac{1}{6}
x_i = a+i\Delta x = -1 + \frac{i}{6}

Thus, for the trapezoidal sum we get:

\sum_{i=1}^{12} \frac{e^{-\left(-1 + i/6\right)^2} + e^{-\left(-1 + (i-1)/6\right)^2}}{2} \frac{1}{6}.

Now, to get a numerical estimate, we can use the code at the bottom of our Riemann sums page. The code is already set up for this integral but the number of terms is different. You can simply edit the code to get

\sum_{i=1}^{12} \frac{e^{-\left(-1 + i/6\right)^2} + e^{-\left(-1 + (i-1)/6\right)^2}}{2} \frac{1}{6} \approx 1.4902.