Question

How many applications of integration by parts are required to evaluate integral (x^7)(e^x) dx?

Answer 1

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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:

$\mathsf{\mathtt{I}_0=\displaystyle\int\! x^k\cdot e^x\,dx\qquad\qquad (k\ge 1,~~k\in\mathbb{N})}$


Now, integrate it by parts:

$\begin{array}{lcl} \mathsf{u=x^k}&\quad\Rightarrow\quad&\mathsf{du=k\cdot x^{k-1}\,dx}\\\\ \mathsf{dv=e^x\,dx}&\quad\Leftarrow\quad&\mathsf{v=e^x} \end{array}$


$\mathsf{\displaystyle\int\! u\,dv=u\cdot v-\int\! v\,du}\\\\\\ \mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-\int\! e^x\cdot k\cdot x^{k-1}\,dx}\\\\\\ \mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-k\int\! x^{k-1}\cdot e^x\,dx}\\\\\\ \mathsf{\displaystyle\int\! x^k\cdot e^x\,dx=x^k\cdot e^x-k\cdot \mathtt{I}_1}$

where $\mathsf{\mathtt{I}_1=\displaystyle\int\! x^{k-1}\cdot e^x\,dx.}$


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:

$\mathsf{\displaystyle\int\! e^x\,dx}$

But this one doesn't need to be evaluated by parts. You can directly write the result:

$\mathsf{\displaystyle\int\! e^x\,dx=e^x+C}$


Shortly, for the integral

$\mathsf{\mathtt{I}_0=\displaystyle\int\! x^7\cdot e^x\,dx}$

you have to apply integration by parts  7  times (not  8  times).


I hope this helps. =)


Tags:  indefinite integral integration by parts reduction formula product polynomial exponential differential integral calculus