2
$\begingroup$

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.

$\endgroup$
5
  • 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.