# What is the average value of the function f(t) = t (sqrt (1 + t^2) ) on the interval [0,5]?

May 13, 2017

$\frac{1}{15} \left(26 \sqrt{26} - 1\right)$

#### Explanation:

The average value of the function $f \left(t\right)$ on $t \in \left[a . b\right]$ is given by:

$\frac{1}{b - a} {\int}_{a}^{b} f \left(t\right) \mathrm{dt}$

So here, the average value of $f \left(t\right) = t \sqrt{1 + {t}^{2}}$ on $t \in \left[0 , 5\right]$ is given by:

$\frac{1}{5 - 0} {\int}_{0}^{5} t \sqrt{1 + {t}^{2}} \mathrm{dt}$

Use the substitution $u = 1 + {t}^{2}$. This implies that $\mathrm{du} = 2 t \mathrm{dt}$. When we change the bounds, $t = 0$ becomes $u = {0}^{2} + 1 = 1$ and $t = 5$ becomes $u = {5}^{2} + 1 = 26$:

$= \frac{1}{5 \times 2} {\int}_{0}^{5} \sqrt{1 + {t}^{2}} \left(2 t \mathrm{dt}\right) = \frac{1}{10} {\int}_{1}^{26} \sqrt{u} \mathrm{du} = \frac{1}{10} {\int}_{1}^{26} {u}^{\frac{1}{2}} \mathrm{du}$

Using $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right)$:

$= \frac{1}{10} {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{1}^{26} = \frac{1}{10} \left(\frac{2}{3}\right) {\left[{u}^{\frac{3}{2}}\right]}_{1}^{26} = \frac{1}{15} \left({26}^{\frac{3}{2}} - {1}^{\frac{3}{2}}\right) = \frac{1}{15} \left(26 \sqrt{26} - 1\right)$