This is the first question from Single Variable Calculus 6th Edition Chapter 7, Section 2. (Authors: Hughes-Hallet, Gleason, McCallum, et al.)

I've formatted the title of this question to be in a more general form, but the question from the book is as follows:

Use integration by parts to express $\int x^2e^xdx$ in terms of: $$ a. \int x^3e^xdx \qquad \qquad b. \int xe^xdx$$

The book also provides the solution in the back, but not the arithmetic to justify it.

  1. (a) $ \frac {x^3e^x}{3} - \frac {1}{3} \int x^3e^xdx $

    (b) $ x^2e^x-2 \int xe^xdx$

What I understand:

Integration by parts.

What I do not understand:

How they got that solution.

  • 2
    $\begingroup$ Hint: apply parts with $u=x^2, dv=e^xdx$. Then apply parts with $u=e^x, dv=x^2dx$. $\endgroup$ – lulu Sep 21 '19 at 22:50
  • $\begingroup$ Just to clarify, the book is basically asking that the anti-derivative of g(x) be constructed using u and dv from f(x)? And what comes before the anti-deriviative of g(x) as a result of integration by parts using f(x) doesn't matter? Thank you for the hint by the way, it helped a lot. $\endgroup$ – Evilscaught Sep 21 '19 at 23:28
  • $\begingroup$ Well, I'm not sure of your notation. The version of parts I am using is $\int u\,dv=uv-\int v \,du$. $\endgroup$ – lulu Sep 21 '19 at 23:32
  • $\begingroup$ The integration by parts notation I'm using is from the book, it is as follows: $ \int uv'dx = uv - \int u'vdx $ But, I have previously seen the notation that you're using. $\endgroup$ – Evilscaught Sep 21 '19 at 23:37
  • $\begingroup$ Well, that's more or less the same...as $dv=v'\,dx$ and $du=u'\,dx$. $\endgroup$ – lulu Sep 21 '19 at 23:39

Integration by parts just exploits the product rule for derivatives and the fundamental theorem of calculus: $fg=\int (fg)'=\int f'g+\int fg'\implies \int f'g=fg-\int fg'$.

So, $(\dfrac{x^3e^x}3)'=x^2e^x+\dfrac {x^3e^x}3\implies \int x^2e^x=\dfrac {x^3e^x}3-\int\dfrac {x^3e^x}3$.

$b)$ is done the same way.


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