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Evaluate using integration by parts ​

Evaluate using integration by parts ​-example-1
User Zoinky
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1 Answer

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Rather than carrying out IBP several times, let's establish a more general result. Let


I(n)=\displaystyle\int x^ne^x\,\mathrm dx

One round of IBP, setting


u=x^n\implies\mathrm du=nx^(n-1)\,\mathrm dx


\mathrm dv=e^x\,\mathrm dx\implies v=e^x

gives


\displaystyle I(n)=x^ne^x-n\int x^(n-1)e^x\,\mathrm dx


I(n)=x^ne^x-nI(n-1)

This is called a power-reduction formula. We could try solving for
I(n) explicitly, but no need.
n=5 is small enough to just expand
I(5) as much as we need to.


I(5)=x^5e^x-5I(4)


I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)


I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)


I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)


I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))

Finally,


I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C

so we end up with


I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C


I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C

and the antiderivative is


\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C

User Edsamiracle
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