Question

Evaluate the definite integral ${\int }_0^1(x{e}^x+\sin \frac{\pi x}{4})dx$

Answer 1

Let $I={\int }_0^1(x{e}^x+\sin \frac{\pi x}{4})dx$
$\Rightarrow \int (x{e}^x+\sin \frac{\pi x}{4})dx=x\int e^{x}dx-\int \{\left(\frac{d}{dx}x\right)\int e^{x}dx\}dx+\left\{\frac{-\cos \frac{\pi x}{4}}{\frac{\pi }{4}}\right\}$
$=x{e}^x-\int e^{x}dx-\frac{4}{\pi }\cos \frac{\pi x}{4}$
$=x{e}^x-e^x-\frac{4}{\pi }\cos \frac{\pi x}{4}=F\left(x\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(1\right)-F\left(0\right)$
$=(1.e^1-e^1-\frac{4}{\pi }\cos \frac{\pi }{4})-(0.e^0-e^0-\frac{4}{\pi }\cos 0)$
$=e-e-\frac{4}{\pi }\left(\frac{1}{\sqrt{2}}\right)+1+\frac{4}{\pi }$
$=1+\frac{4}{\pi }-\frac{2\sqrt{2}}{\pi }$