109k views
4 votes
Check to see if I’m starting good and help to end

Check to see if I’m starting good and help to end-example-1

1 Answer

7 votes

For this problem we have a certain function,


MP(x)=xe^(-0.6x)

And we need to integrate it over the interal:


0\le x\le100

In order to solve this problem, we need to apply the integration by parts method. Such as:


\int udv=uv-\int vdu

For our case we will call "u = x", therfore we have:


\begin{gathered} u=x \\ (du)/(dx)=1 \\ du=dx \end{gathered}

And the other part of the integral will be dv, we have:


\begin{gathered} e^(-0.6x)dx=dv \\ \int e^(-0.6x)dx=\int dv \\ -1.67e^(-0.6x)=v \\ e^(-0.6x)=-(v)/(1.67) \\ e^(-0.6x)=-0.6v \end{gathered}

Using the second expression, we have:


\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}

Now we need to apply the interval, to define the function:


\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}

The value for the integral on the interval from 0 to 100 is 2.78.

User Varma
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories