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Prove the following\[\int\limits_0^1x\;e^x\;dx=1\]

1 Answer

  • (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,
on applying the limits we get
But $ e^0=1$
Therefore $1e^1-e^1+1$
Therefore I=1.
Hence proved


answered Feb 19, 2013 by meena.p