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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Prove the following\[\int\limits_0^1x\;e^x\;dx=1\]

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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 meena.p
 
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