Question

Find the values of $c$ that satisfy the MVT for integrals on $[0,1]$.
$f\left(x\right)=x(1-x)$

• A. $c=\frac{1}{2}$
• B. $c=-\frac{1}{2}$
• C. $c=\frac{2}{3}$
• D. $c=-\frac{1}{3}$

Answer 1

The correct option is A $c=\frac{1}{2}$

Given $f\left(x\right)=x(1-x)$ and $a=0$ and $b=1$.

So $f\left(a\right)=0$ and $f\left(b\right)=0$.

Now by MVT:

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$=>(1-c)-c=0$

$=>1-2c=0$

hence, $c=\frac{1}{2}$