Answer
(b) $-7$
(a) $\frac{4}{3}$
Work Step by Step
Mean-Value Theorem for Integrals: $\int_{a}^{b} f(x) d x=f\left(x^{*}\right)(-a+b)$
(a) $\int_{0}^{3} \sqrt{x} d x=2 \sqrt{3}$
$\Rightarrow 2 \sqrt{3}=f\left(x^{*}\right).3$
$\Rightarrow \frac{2 \sqrt{3}}{3}=f\left(x^{*}\right)=\sqrt{x^{*}}$
$\Rightarrow x^{*}=\frac{4 \cdot 3}{9}=\frac{4}{3}$
(b) $\int_{-12}^{0}\left(x+x^{2}\right) d x=504$
$\Rightarrow 504=f\left(x^{*}\right).12$
$\Rightarrow 42=f\left(x^{*}\right)=\left(x^{*}\right)^{2}+x^{*}$
$\Rightarrow\left(x^{*}\right)^{2}-42+x^{*}=0$
$\Rightarrow x^{*}=-7$ or $x^{*}=6$
$\Rightarrow x^{*}=-7$ since $x^{*}$ has to lie between -12, 0.
