Question

Let $f\left(a\right)>0$, and let$f\left(x\right)$ 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 $f\left(b\right)$ and minimum value $f\left(a\right)$.
• B. maximum value $bf\left(b\right)$ and minimum value $af\left(a\right)$.
• C. maximum value $f\left(a\right)$ and minimum value $f\left(b\right)$.
• D. none of these

Answer 1

The correct option is A maximum value $f\left(b\right)$ and minimum value $f\left(a\right)$.

It is given that $f\left(x\right)$ is a non-decreasing function on $[a,b]$.
$\therefore f\left(a\right)\le f\left(x\right)\le f\left(b\right);\forall x\in [a,b]$
$\Rightarrow {\int }_a^{b}f\left(a\right)dx\le {\int }_a^{b}f\left(x\right)dx\le {\int }_a^{b}f\left(b\right)dx$
$\Rightarrow (b-a)f\left(a\right)\le {\int }_a^{b}f\left(x\right)dx\le (b-a)f\left(b\right)$
$\Rightarrow f\left(a\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le f\left(b\right)$

Hence, option A.