Answer:
[tex]f_{avg}=\frac{1}{8}(5e^4-1)[/tex]
Step-by-step explanation:
We are given that a function
[tex]f(x)=x^2e^{2x}[/tex]
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
[tex]\frac{1}{b-a}\int_{a}^{b}f(x)dx[/tex]
Using the formula
[tex]f_{avg}=\frac{1}{2-0}\int_{0}^{2}x^2e^{2x} dx[/tex]
By Parts integration formula
[tex]\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx[/tex]
u=[tex]x^2[/tex] and v=[tex]e^{2x}[/tex]
Apply by parts integration
[tex]f_{avg}=\frac{1}{2-0}([\frac{x^2e^{2x}}{2}]^{2}_{0}-\int_{0}^{2}(2x\times \frac{e^{2x}}{2}dx)[/tex]
[tex]f_{avg}=\frac{1}{2}(2e^4-(\int_{0}^{2}xe^{2x})dx[/tex]
[tex]f_{avg}=\frac{1}{2}(2e^4-0-([\frac{xe^{2x}}{2}]^{2}-\frac{1}{4}[e^{2x}]^{2}_{0}))[/tex]
[tex]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})[/tex]
[tex]f_{avg}=\frac{1}{8}(5e^4-1)[/tex]
