Browse Questions

# Prove the following$\int\limits_0^1x\;e^x\;dx=1$

Can you answer this question?

Toolbox:
• (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
• (ii)$\int udv=uv-\int vdu$
• (iii)$\int e^x dx=e^x dx$
Given $I=\int\limits_0^1x\;e^x\;dx=1$

Clearly this is of the form $\int udv=uv-\int vdu$

Let $u=x$ on differentiating w.r.t x we get, $du=dx$

Let $dv=e^x.dx4$ on integrating we get,$v=e^x$

Now substituting for u,v,du and dv we get,

$I=\int\limits_0^4x\;e^x\;dx=(xe^x)_0^1-\int _0^1 e^x.dx$

On integrating we get,

$I=[xe^x]_0^1-[e^x]_0^1$

on applying the limits we get

$(1e^1-0)-[e^1-e^0]$

But $e^0=1$

Therefore $1e^1-e^1+1$

$=1$

Therefore I=1.

Hence proved

answered Feb 19, 2013 by