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Make it more general, and see what happens when you try to reduce the exponent of
x in the following integral:
![\mathsf{\mathtt{I}_0=\displaystyle\int\! x^k\cdot e^x\,dx\qquad\qquad (k\ge 1,~~k\in\mathbb{N})}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cmathtt%7BI%7D_0%3D%5Cdisplaystyle%5Cint%5C%21%20x%5Ek%5Ccdot%20e%5Ex%5C%2Cdx%5Cqquad%5Cqquad%20%28k%5Cge%201%2C~~k%5Cin%5Cmathbb%7BN%7D%29%7D)
Now, integrate it by parts:
![\begin{array}{lcl} \mathsf{u=x^k}&\quad\Rightarrow\quad&\mathsf{du=k\cdot x^{k-1}\,dx}\\\\ \mathsf{dv=e^x\,dx}&\quad\Leftarrow\quad&\mathsf{v=e^x} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcl%7D%0A%5Cmathsf%7Bu%3Dx%5Ek%7D%26%5Cquad%5CRightarrow%5Cquad%26%5Cmathsf%7Bdu%3Dk%5Ccdot%20x%5E%7Bk-1%7D%5C%2Cdx%7D%5C%5C%5C%5C%0A%5Cmathsf%7Bdv%3De%5Ex%5C%2Cdx%7D%26%5Cquad%5CLeftarrow%5Cquad%26%5Cmathsf%7Bv%3De%5Ex%7D%0A%5Cend%7Barray%7D)
![\mathsf{\displaystyle\int\! u\,dv=u\cdot v-\int\! v\,du}\\\\\\ \mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-\int\! e^x\cdot k\cdot x^{k-1}\,dx}\\\\\\ \mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-k\int\! x^{k-1}\cdot e^x\,dx}\\\\\\ \mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-k\cdot \mathtt{I}_1}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20u%5C%2Cdv%3Du%5Ccdot%20v-%5Cint%5C%21%20v%5C%2Cdu%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20x%5Ek%5Ccdot%20e%5Ex%5C%2Cdx%3Dx%5Ek%5Ccdot%20e%5Ex-%5Cint%5C%21%20e%5Ex%5Ccdot%20k%5Ccdot%20x%5E%7Bk-1%7D%5C%2Cdx%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20x%5Ek%5Ccdot%20e%5Ex%5C%2Cdx%3Dx%5Ek%5Ccdot%20e%5Ex-k%5Cint%5C%21%20x%5E%7Bk-1%7D%5Ccdot%20e%5Ex%5C%2Cdx%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20x%5Ek%5Ccdot%20e%5Ex%5C%2Cdx%3Dx%5Ek%5Ccdot%20e%5Ex-k%5Ccdot%20%5Cmathtt%7BI%7D_1%7D)
where
![\mathsf{\mathtt{I}_1=\displaystyle\int\! x^{k-1}\cdot e^x\,dx.}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cmathtt%7BI%7D_1%3D%5Cdisplaystyle%5Cint%5C%21%20x%5E%7Bk-1%7D%5Ccdot%20e%5Ex%5C%2Cdx.%7D)
So after one iteration, the exponent of
x was decreased by one unit.
The question is:
after how many iterations will the exponent of x equals zero? After exactly
k iterations, of course.
Therefore, for
k = 7, you have to apply integration by parts
7 times, to get rid of that polynomial factor. Then, there will be one last integral left to evaluate:
![\mathsf{\displaystyle\int\! e^x\,dx}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20e%5Ex%5C%2Cdx%7D)
But this one doesn't need to be evaluated by parts. You can directly write the result:
![\mathsf{\displaystyle\int\! e^x\,dx=e^x+C}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20e%5Ex%5C%2Cdx%3De%5Ex%2BC%7D)
Shortly, for the integral
![\mathsf{\mathtt{I}_0=\displaystyle\int\! x^7\cdot e^x\,dx}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cmathtt%7BI%7D_0%3D%5Cdisplaystyle%5Cint%5C%21%20x%5E7%5Ccdot%20e%5Ex%5C%2Cdx%7D)
you have to apply integration by parts
7 times (not
8 times).
I hope this helps. =)
Tags: <em>indefinite integral integration by parts reduction formula product polynomial exponential differential integral calculus</em>