Question

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

Answer 1

${\int }_0^1[x{e}^x+sin(\frac{\pi x}{4}\left)\right]dx=[x{e}^x{]}_0^1-{\int }_0^{1}1.e^{x}dx-{\left[\right(\frac{cos\left(\frac{\pi x}{4}\right)}{\left(\frac{\pi }{4}\right)}\left)\right]}_0^1=(e-0)-[e^x{]}_0^1-\left(\frac{4}{\pi }\right)\left(\right(\frac{1}{\sqrt{2}})-1)=e-(e-1)-\left(\frac{2\sqrt{2}}{\pi }\right)+\left(\frac{4}{\pi }\right)=\left(\frac{4-2\sqrt{2}}{\pi }\right)+1$