CPM Homework Help : APCALC Problem 8-74

Multiple Choice: The average value of $f(x)=\ln(x^2)$ on the interval $1\le x\le e$ is closest to:

$0.50$

$0.65$

$0.75$

$0.85$

$1.00$

Hint:

You are given $f(x)$ and you want to find the average (mean) value of $f(x)$ . Will you integrate and divide, or will you find the slope of the secant line?

Average (Mean) Values

To calculate the mean (average) value of a finite set of items, add up the values of items and divide by the number of items.

Integrals help us add over a continuous interval. Therefore, for any continuous function $f$ :

$\frac{\int_a^bf(x)dx}{b-a}=$ mean value of $f$ over $[a, b]$

First quadrant, bell curve labeled, f of x, left end point on the y axis, labeled, a, right end point labeled, b, dashed horizontal segment, about 1 fourth up from x axis to peak, labeled average, & shaded rectangle between, A & b, segment & x axis.

Since $\int _ { a } ^ { b } f ( x ) d x$ , we can also calculate the average value of any function $f$ using its antiderivative $F$ . Its average slope gives the average rate of change of $F$ , which is the same as the average value of $f.$

$\frac{\int_a^bf(x)dx}{b-a}=\frac { F ( b ) - F ( a ) } { b - a }=$ mean rate of change of $F ′$ over $[a, b]$

First quadrant, 2 tick marks on x axis, first at the origin labeled, A, second almost to the right end, labeled b, Increasing curve labeled, capital F of x, starting at the origin, changing from concave up to concave down, in center of quadrant, ending at point corresponding to, b, almost at the top, with dashed segment labeled, m = average, from origin to end point of curve.