and v=
![f_{avg}=\frac{1}{2-0}([\frac{x^2e^{2x}}{2}]^{2}_{0}-\int_{0}^{2}(2x\times \frac{e^{2x}}{2}dx)](/tpl/images/0455/1845/06463.png)
![f_{avg}=\frac{1}{2}(2e^4-0-([\frac{xe^{2x}}{2}]^{2}-\frac{1}{4}[e^{2x}]^{2}_{0}))](/tpl/images/0455/1845/da1d5.png)
Find the average value of the function on the given interval.
f(x)=x^2e^2x; [0,2]...
Mathematics, 15.01.2020 00:31 gabedafame26
Find the average value of the function on the given interval.
f(x)=x^2e^2x; [0,2]
Answers: 2
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