Chapter 5.4, Problem 68E

Bounds on an integral Suppose f is continuous on [a, b] with f″(x) > 0 on the interval. It can be shown that

$(b-a)f\left(\frac{a+b}{2}\right)\le {\displaystyle {\int }_a^{b}f(x)dx\le (b-a)\frac{f(a)+f(b)}{2}.}$

1. a. Assuming f is nonnegative on [a, b], draw a figure to illustrate the geometric meaning of these inequalities. Discuss your conclusions.
2. b. Divide these inequalities by (b − a) and interpret the resulting inequalities in terms of the average value of f on [a, b].