Sunday, August 21, 2016

`int z^3 e ^ z dz` Evaluate the integral

`intz^3e^zdz`


If f(x) and g(x) are differentiable functions, then


`intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx`


If we write f(x)=u and g'(x)=v, then


`intuvdx=uintvdx-int(u'intvdx)dx`


Using the above integration by parts,


Let `u=z^3 , u'=3z^2`


and let `v=e^z, v'=e^z`


`intz^3e^z=z^3inte^zdz-int(3z^2inte^zdz)dz`


`=z^3e^z-int(3z^2e^z)dz`


`=z^3e^z-3intz^2e^zdz`


again applying integration by parts,


`=z^3e^z-3(z^2inte^zdz-int(d/dz(z^2)inte^zdz)dz`


`=z^3e^z-3(z^2e^z-int(2ze^z)dz`


`=z^3e^z-3z^2e^z+6intze^zdz`


again applying integration by parts,


`=z^3e^z-3z^2e^z+6(zinte^zdz-int(d/dz(z)inte^zdz)dz)`


`=z^3e^z-3z^2e^z+6(ze^z-int(1*e^z)dz)`


`=z^3e^z-3z^2e^z+6(ze^z-e^z)`


adding constant to the solution,


`=z^3e^z-3z^2e^z+6ze^z-6e^z+C`

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