Question #7bb57

Nov 11, 2016

Solve $\sqrt{x + 1} = \frac{1}{3} {\int}_{0}^{3} \sqrt{x + 1} \mathrm{dx}$

Explanation:

The mean value theorem for integrals says that if $f$ is continuous on $\left[a , b\right]$ then there is a $c$ in $\left(a , b\right)$ such that $f \left(c\right)$ equals the average value of $f$ on $\left[a , b\right]$.

To find these $c$'s, solve the equation

$f \left(x\right) = \frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$. Retain only solutions in $\left(a , b\right)$.

In this question, we get

$\sqrt{x + 1} = \frac{1}{3} {\left.{\left(x + 1\right)}^{\frac{3}{2}}\right]}_{0}^{3} = \frac{7}{3}$

So $x = \frac{40}{9}$