Question

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

Answer 1

${\int }_0^1(x{e}^x+sin\frac{\pi .x}{4})dx={\int }_0^{1}x{e}^{x}dx+{\int }_0^{1}sin\left(\frac{\pi .x}{4}\right)dx=[x{e}^x-\int 1.e^{x}dx{]}_0^1+{\left[\frac{-cos\frac{\pi .x}{4}}{\frac{\pi }{4}}\right]}_0^1$
First integral solved by using the integration by parts and for second $(\int sinxdx=-\frac{cosax}{a})$
$=[x{e}^x-e^x{]}_0^1-\frac{4}{\pi }{\left[cos\right(\frac{\pi .x}{4}\left)\right]}_0^1=(e-e)-(0-e^0)-\frac{4}{\pi }(cos\frac{\pi }{4}-cos0)=0-(-1)-\frac{4}{\pi }(\frac{1}{\sqrt{2}}-1)=1-\frac{4}{\pi }\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)\times \frac{\sqrt{2}}{\sqrt{2}}=1+\frac{2}{\pi }(2-\sqrt{2})=1+\frac{4}{\pi }-\frac{2\sqrt{2}}{\pi }.$