# What is the average value of the function f(x) = 2x^3(1+x^2)^4 on the interval [0,2]?

##### 1 Answer
Apr 11, 2016

The average value is $\frac{4948}{5} = 989.6$

#### Explanation:

The average value of $f$ on interval $\left[a , b\right]$ is $\frac{1}{b - a} {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

So we get:

$\frac{1}{2 - 0} {\int}_{0}^{2} 2 {x}^{3} {\left({x}^{2} + 1\right)}^{4} \mathrm{dx} = \frac{2}{2} {\int}_{0}^{2} {x}^{3} \left({x}^{8} + 4 {x}^{6} + 10 {x}^{4} + 4 {x}^{2} + 1\right) \mathrm{dx}$

$= {\int}_{0}^{2} \left({x}^{11} + 4 {x}^{9} + 10 {x}^{7} + 4 {x}^{5} + {x}^{3}\right) \mathrm{dx}$

 = x^12/12+(4x^10)/10 + (6x^8)/8+(4x^6)/6+x^4/4]_0^2

$= {\left(2\right)}^{12} / 12 + \frac{2 {\left(2\right)}^{10}}{5} + \frac{3 {\left(2\right)}^{8}}{4} + \frac{2 {\left(2\right)}^{6}}{3} + {\left(2\right)}^{4} / 4$

$= \frac{4948}{5} = \frac{9896}{10} = 989.6$