The Height With the Same Area

You'll learnto read the average value as the height of the equal-area rectangle, to divide by the length of the interval rather than by anything else, to find where the function attains that height, and to check an answer against the values the function actually takes.

An integral is an area. An average value is a height. One division separates them, and it is the step that goes missing more often than any other in this unit. Everything here follows from asking a single question: what constant height would enclose the same area over the same base? Every common error in this topic turns out to be a wrong divisor, and two of the four land close enough to the right answer to survive unnoticed.

8 STEPS · 6 QUICK CHECKS · DIVIDE BY THE LENGTH, THEN CHECK IT IS A VALUE OF f · v1

f_avg = (1/(b − a)) ∫ f dx ⇒ the height of the equal-area rectangle ⇒ a value of f, not an area

Before you start

What you're looking at

A curve, the region under it, and a rectangle standing on the same base. The rectangle's height is adjusted until the two enclose the same area — and that height is the average value.

The question

Given the area, what height would produce it over this base, and is the number you get something the function could actually equal?

Watch for

The rectangle's base is the length of the interval, not the right endpoint. On an interval starting at zero those two are the same number, which is why the error hides there.

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