Question

\[\int\limits_0^1 \left( x e^{2x} + \sin\frac{\ pix}{2} \right) dx\]

Answer 1

\[Let I = \int_0^1 \left( x e^{2x} + \sin \frac{\ pix}{2} \right) d x . Then, \]
\[I = \int_0^1 x e^{2x} d x + \int_0^1 \sin \frac{\ pix}{2} dx\]
\[\text{Integrating first term by parts}\]
\[I = \left[ x \frac{e^{2x}}{2} \right]_0^1 - \int_0^1 1 \frac{e^{2x}}{2} dx + \left[ - \frac{\cos \frac{\ pix}{2}}{\frac{\pi}{2}} \right]_0^1 \]
\[ \Rightarrow I = \left[ x \frac{e^{2x}}{2} \right]_0^1 - \left[ \frac{e^{2x}}{4} \right]_0^1 - \frac{2}{\pi} \left[ \cos \frac{\ pix}{2} \right]_0^1 \]
\[ \Rightarrow I = \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} + \frac{2}{\pi}\]
\[ \Rightarrow I = \frac{e^2}{4} + \frac{1}{4} + \frac{2}{\pi}\]