Question

If $f$ satisfies $|f\left(u\right)-f\left(\nu \right)|\le |u-\nu |$ for $u,\nu \in [a,b]$,

then the maximum value of $\left|{\int }_a^{b}f\right(x)dx-(b-a\left)f\right(a\left)\right|$ is?

• A. $\frac{b-a}{2}$
• B. $\frac{(b-a{)}^2}{2}$
• C. $(b-a{)}^2$
• D. None of these

Answer 1

Answer: (B)

$\frac{(b-a{)}^2}{2}$

$\left|{\int }_a^{b}f\right(x)dx-(b-a\left)f\right(a\left)\right|$
$=\left|{\int }_a^{b}f\right(x)dx-{\int }_a^{b}f(a\left)dx\right|=\left|{\int }_a^b\left[f\right(x)-f(a\left)\right]dx\right|$

$\Rightarrow \left|{\int }_a^{b}f\right(x)dx-(b-a\left)f\right(a\left)\right|$$\equiv {\int }_a^b\left|f\right(x)-f(a\left)\right|dx$

Using

$|f\left(u\right)-f\left(\nu \right)|\le |u-\nu |$

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

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

Hence, option B.