How do you find the average value of a function f(x)=(x−1)2 on the interval from x=1 to x=5?

Question

Kapovkah7hpl · Accepted Answer

The question asks for the average value:
The average value of a function over an interval is the height a rectangle on that interval would need to have, if we want the area of the rectangle to be equal to the area under the graph of the function.
To find that height, find the area under the graph, then divide by the length of the interval. (Or multiply by 1 over the length of the interval.)
Average value

= \frac{1}{5 - 1} \int_{1}^{5} {(x - 1)}^{2} d x = \frac{1}{4} \int_{1}^{5} {(x - 1)}^{2} d x

We could evaluate the integral by expanding and evaluating the resulting quadratic expression, but I think this one is cleaner is we just substitute u=x-1

\frac{1}{4} {[\frac{{(x - 1)}^{3}}{3}]}_{1}^{5} = \frac{1}{4} [\frac{4^{3}}{3} - \frac{0}{3}] = \frac{16}{3}

Note Although the question asks for average value, it was posted under "Average rate of change" which would be:

\frac{△ y}{△ x} = \frac{f (5) - f (1)}{5 - 1} = \frac{16 - 0}{4} = 4

How do you find the average value of

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