# Evaluate the definite integral$\int\limits_0^1(xe^x+\sin\frac{\pi\;x}{4})dx$

$\begin{array}{1 1} 1+\large \frac{4}{\pi}-\frac{2 \sqrt 2}{\pi} \\1-\large \frac{4}{\pi}+\frac{2 \sqrt 2}{\pi} \\ 1-\large \frac{2}{\pi}+\frac{4 \sqrt 2}{\pi} \\ 1-\large \frac{2}{\pi}-\frac{2 \sqrt 4}{\pi} \end{array}$

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-f(a)$
• (ii) Methods of integration by parts: $\int udv=uv-\int vdu$
• (iii)$\int \sin x dx=-\cos x+c$
Given $\int\limits_0^1(xe^x+\sin\frac{\pi\;x}{4})dx$

On seperating the terms

$I=\int \limits_0^1 xe^xdx+\int\limits_0^1\sin \frac{\pi x}{4}dx$

Consider $I_1=\int \limits_0^1 x e^xdx$

This is of the form $\int udv$

Where $\int udv=uv-\int vdu$

Let u=x

On differentiating with respect to x,

du=dx

Let $dv=e^xdx$

On integrating we get,

$v=e^x$

$\int \limits_0^1 xe^x=(xe^x)_0^1-\int \limits_0^1e^x dx$

On integrating we get

$I_1=(xe^x)_0^1-(e^x)_0^1$

Consider $I_2=\int \limits_0^1 \sin \frac{\pi}{4}x$

$=\frac{1}{\pi/4}(\cos \pi/4 x)_0^1$

$I=I_1+I_2$

$=(xe^x)_0^1-(e^x)_0^1+(-\frac{4}{\pi} \cos \frac{\pi}{4}x)_0^1$

On applying limits

$[1.e^1-0]-[e^1-e^0]+[(-\frac{4}{\pi}.\cos \frac{\pi}{4}-(-\frac{4}{\pi} \cos 0)]$

$e^0=1;\cos \frac{\pi}{4}=\frac{1}{\sqrt 2}; \cos 0=1$

Therefoer $I=1-\frac{4}{\pi}.\frac{1}{\sqrt 2}+\frac{4}{\pi}$

$\int \limits _0^1 (xe^x+\sin \large\frac{\pi x}{4})dx=1+\frac{4}{\pi}-\frac{2 \sqrt 2}{\pi}$