# Evaluate the definite integral$\int\limits_0^1x\;e^{x^2}dx$

Toolbox:
• (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
• (ii)If there are two functions u and v, and the integral function is of the form $\int udv,$then it can be solved by the method of integration by parts.$\int udv=uv-\int vdu$
• (iii)$\int e^x=e^x+c.$
Given $\int\limits_0^1x\;e^{x^2}dx$

Let $x^2=t$

On differentiating we get w.r.t.x

$2xdx=dt$

Therefore $xdx=dt/2$

Substituting $t_1$ and dt we get

$I=\frac{1}{2}\int \limits_0^1 e^t dt$

On integrating we get,

$\frac{1}{2}[e^t]_0^1$

$=\frac{1}{2}[e^1-e^0]$

But $e^0=1$

$I=\frac{1}{2}[e-1]$

$\int \limits_0^1 xe^xdx=\frac{1}{2}[e-1]$