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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
