Question

Let $f\left(a\right)>0$, and $f\left(x\right)$ be a non-decreasing continuous function in $[a,b]$. Then $\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$ has the

• A. maximum value of $f\left(b\right)$
• B. minimum value of $f\left(a\right)$
• C. maximum value of $bf\left(b\right)$
• D. minimum value $\frac{f\left(a\right)}{b-a}$

Answer 1

Answer: (A), (B)

minimum value of $f\left(a\right)$
maximum value of $f\left(b\right)$

$f\left(x\right)$ is a non-decreasing function in $[a,b]$.
Hence, for $a\le x\le b$ , $f\left(a\right)\le f\left(x\right)\le f\left(b\right)$
$\frac{1}{b-a}{\int }_a^{b}f\left(a\right)dx\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le \frac{1}{b-a}{\int }_a^{b}f\left(b\right)dx$
$\Rightarrow f\left(a\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le f\left(b\right)$