Applications of integrals, part 1: Areas & Average Value

Usually the first application of integration is to find the area bounded by a function and the x-axis, followed by finding the area between two functions. We begin with these problems

First some calculator hints

Graphing Integrals using a graphing calculator to graph functions defined by integrals

Graphing Calculator Use  and Definition Integrals – Exam considerations Suggestions for using a calculator efficiently in area/volume problems

Area Problems

Area Between Curves

Under is a Long Way Down How to avoid “negative area.”

Density Functions Not often asked on the AP exams, but a good application related to area, nevertheless.

Who’d a thunk it? Some more complicated area problems for CAS solution.

Improper Integrals and Proper Areas – a BC topic

Average Value

Average Value of a Function

What’s a Mean Old Average Anyway – Discusses the different “average” in calculus

Half-full and Half Empty – Average Value

Average Value Activity to help students discover the Average Value formula








Area Between Curves

Applications of Integration 1 – Area Between Curves

The first thing to keep in mind when teaching the applications of integration is Riemann sums. The thing is that when you set up and solve the majority of application problems you cannot help but develop a formula for the situation. Students think formulas are handy and go about memorizing them badly. By which I mean they forget or never learn where the various things in the formula come from. A slight change in the situation and they are lost. Behind every definite integral stands a Riemann sum; each application should be approached through its Riemann sum. If students understand that, they will make fewer mistakes with the “formula.”

As I suggested in a previous post, I believe all area problems should be treated as the area between two curves. If you build the Riemann sum rectangle between the graph and the axis and calculate its vertical side as the upper function minus the lower (or right minus left if you use horizontal rectangles) you will always get the correct integral for the area. If the upper curve is the x-axis, then the vertical sides of the Riemann sums are (0 – f(x)) and you get a positive area as you should.

If both your curves are above the x-axis, then it is tempting to explain what you are doing as subtracting the area between the lower curve and the x-axis from the area between the upper curve and the x-axis. And this is not wrong. It just does not work very smoothly when one, both or parts of either are below the x-axis. Then you go into all kinds of contortions explaining things in terms of positive and negative areas.  Why go there?

Regardless of where the two curves are relative to the x-axis, the vertical distance between them is the upper value minus the lower, f(x) – g(x). It does not matter if one or both functions are negative on all or part of the interval, the difference is positive and the area between them is

\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\left( f\left( {{x}_{i}} \right)-g\left( {{x}_{i}} \right) \right)\Delta {{x}_{i}}}=\int_{a}^{b}{f\left( x \right)-g\left( x \right)dx}.

Furthermore, this Riemann sum rectangle is used in other applications. It is the one rotated in both the washer and shell method of finding volumes. So in area and all applications be sure your students don’t just memorize formulas, but keep their eyes on the rectangle and the Riemann sum.

Finally, if the graphs cross in the interval so that the upper and lower curves change place, you may (1) either break the problem into several pieces so that your integrands are always of the form upper minus lower, or (2) if you intend to do the computation using technology, set up the integral as

\displaystyle \int_{a}^{b}{\left| f\left( x \right)-g\left( x \right) \right|dx}.