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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[x^2e^x\]

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Toolbox:
  • (i)Let us consider two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts\[\int udv=uv-\int vdu\]
  • (ii)$\int e^xdx=e^x+c.$
Given $I=\int x^2e^x dx.$
 
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts where \[\int udv=uv-\int vdu\]
 
Let $u=x^2$.
 
On differentiating we get
 
du=2xdx.
 
Let $dv=e^xdx.$
 
On integrating on both sides we get
 
$v=e^x$
 
On substituting for u,v,du and dx we get,
 
$\int x^2e^xdx=(x^2.e^x)-\int e^x.2xdx.$
 
$\;\;\;=(x^3e^x)-2\int e^x.xdx.$-----(1)
 
Again $\int e^xdx $ is of the form $\int u dv.$
 
Let u=x.
 
On differentiating we get
 
du=dx.
 
let $dv=e^xdx.$
 
On integrating we get
 
$v=e^x.$
 
Hence on substituting for u,v,du and dv we get
 
$\int x.e^xdx=(xe^x)-\int e^xdx.$
 
On integrating we get
 
$xe^x-e^x+c$-----(2)
 
Substituting this in equ(1) we get
 
$\int x^2e^xdx=x^2e^x-2(xe^x-e^x)+c.$
 
Taking $e^x$ as the common factor we get
 
$\;\;\;=e^x[x^2-2x+2]+c.$

 

answered Feb 8, 2013 by sreemathi.v
 
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