For this problem we have a certain function,
[tex]MP(x)=xe^{-0.6x}[/tex]
And we need to integrate it over the interal:
[tex]0\le x\le100[/tex]
In order to solve this problem, we need to apply the integration by parts method. Such as:
[tex]\int udv=uv-\int vdu[/tex]
For our case we will call "u = x", therfore we have:
[tex]\begin{gathered} u=x \\ \frac{du}{dx}=1 \\ du=dx \end{gathered}[/tex]
And the other part of the integral will be dv, we have:
[tex]\begin{gathered} e^{-0.6x}dx=dv \\ \int e^{-0.6x}dx=\int dv \\ -1.67e^{-0.6x}=v \\ e^{-0.6x}=-\frac{v}{1.67} \\ e^{-0.6x}=-0.6v \end{gathered}[/tex]
Using the second expression, we have:
[tex]\begin{gathered} \int xe^{-0.6x}=-1.67xe^{-0.6x}+\int 1.67e^{-0.6x}dx \\ \int xe^{-0.6x}=-1.67xe^{-0.6x}-2.78e^{-0.6x}+K \end{gathered}[/tex]
Now we need to apply the interval, to define the function:
[tex]\begin{gathered} \int ^{100}_0xe^{-0.6x}=(-1.67xe^{-0.6x}-2.78e^{-0.6x})\begin{cases}100 \\ 0\end{cases} \\ \int ^{100}_0xe^{-0.6x}=(-1.67\cdot100\cdot e^{-0.6\cdot100}-2.78e^{-0.6\cdot100})-(-1.67\cdot0\cdot e^{-0.6\cdot0}-2.78e^{-0.6\cdot0}) \\ \int ^{100}_0xe^{-0.6x}=-167e^{-60}-2.78e^{-60}+2.78e^0 \\ \int ^{100}_0xe^{-0.6x}=2.78 \end{gathered}[/tex]
The value for the integral on the interval from 0 to 100 is 2.78.
