Question

Find the average value of the function on the given interval. f(x)=x^2e^2x; [0,2]

Answer 1

Answer:

$f_{avg}=\frac{1}{8}(5e^4-1)$

Step-by-step explanation:

We are given that a function

$f(x)=x^2e^{2x}$

We have to find the average value of function on the given interval [0,2]

Average value of function on interval [a,b] is given by

$\frac{1}{b-a}\int_{a}^{b}f(x)dx$

Using the formula  

$f_{avg}=\frac{1}{2-0}\int_{0}^{2}x^2e^{2x} dx$

By Parts integration formula

$\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx$

u=$x^2$ and v=$e^{2x}$

Apply by parts integration

$f_{avg}=\frac{1}{2-0}([\frac{x^2e^{2x}}{2}]^{2}_{0}-\int_{0}^{2}(2x\times \frac{e^{2x}}{2}dx)$

$f_{avg}=\frac{1}{2}(2e^4-(\int_{0}^{2}xe^{2x})dx$

$f_{avg}=\frac{1}{2}(2e^4-0-([\frac{xe^{2x}}{2}]^{2}-\frac{1}{4}[e^{2x}]^{2}_{0}))$

$f_{avg}=\frac{1}{2}(2e^4-e^4+\frac{1}{4}(e^4-1))=\frac{1}{2}(e^4+\frac{1}{4}e^4-\frac{1}{4})$

$f_{avg}=\frac{1}{8}(5e^4-1)$