Science:Math Exam Resources/Courses/MATH200/April 2012/Question 06 (b)/Solution 1

Using our answer from part (a), we can write I with the given ƒ as

$I=\int _{-1}^{2}\int _{x^{2}}^{x+2}{\frac {\exp(x)}{2-x}}\,dydx.$

We can evaluate the integral as follows:

${\begin{aligned}I&=\int _{-1}^{2}\int _{x^{2}}^{x+2}{\frac {\exp(x)}{2-x}}\,dydx\\&=\int _{-1}^{2}{\frac {\exp(x)}{2-x}}\left(x+2-x^{2}\right)\,dx\\&=\int _{-1}^{2}{\frac {\exp(x)}{2-x}}(-x+2)(x+1)\,dx\\&=\int _{-1}^{2}\exp(x)(x+1)\,dx.\end{aligned}}$

Using integration by parts, with u=x+1 and dv = exp(x),

${\begin{aligned}I&=\int _{-1}^{2}\exp(x)(x+1)\,dx\\&=\exp(x)(x+1){\Big \vert }_{-1}^{2}-\int _{-1}^{2}\exp(x)\,dx\\&=3\exp(2)-\left(\exp(x){\Big \vert }_{-1}^{2}\right)=2\exp(2)+\exp(-1)\end{aligned}}$

Therefore,

${\color {blue}\int _{-1}^{2}\int _{x^{2}}^{x+2}{\frac {\exp(x)}{2-x}}\,dydx=2e^{2}+e^{-1}}$