Use integration to solve for value from 4 to 2
xe^(4-x)
Could someone show me how to do this question I know its by parts but it get lost and end up with an enormous answer, stage by stage would be a great help thank you
Hello, pip1690!
$\displaystyle \int^4_2 xe^{4-x}\,dx$
. . $\displaystyle \begin{array}{ccccccc}u &=& x &\quad & dv &=&e^{4-x}dx \\
du &=& dx & \quad & v &=& -e^{4-x} \end{array}$
We have: .$\displaystyle -xe^{4-x} + \int^4_2e^{4-x}dx \;\;=\;\;-xe^{4-x} - e^{4-x}\,\bigg]^4_2 \;\;=\;\;-e^{4-x}(x+1)\,\bigg]^4_2$
. . $\displaystyle =\;\;\bigg[-e^0(5)\bigg] - \bigg[-e^2(3)\bigg]\;\;=\;\;-5 + 3e^2 $