0
$\begingroup$

I encountered the following question in my book:

"Integrate $f(x)=\sqrt{1+x^2}$ with respect to $x^2$."

I am a bit confused about what this is supposed to mean.

In general, what does it mean to integrate a function $f(x)$ with respect to a function $g(x)$?

$\endgroup$
  • 2
    $\begingroup$ My guess is the Riemann–Stieltjes integral. $\endgroup$ – Harald Hanche-Olsen Oct 25 '12 at 10:21
  • $\begingroup$ Probably that book has said what it means at some previous place... $\endgroup$ – GEdgar Oct 25 '12 at 11:50
  • $\begingroup$ Off-topic note. There are 51 users here with username "Chris" or "chris". $\endgroup$ – GEdgar Oct 25 '12 at 11:52
0
$\begingroup$

In my opinion you are working with Stieltjes integration. See here for a detailed introduction and examples:

http://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral

http://en.wikipedia.org/wiki/Lebesgue%E2%80%93Stieltjes_integration

http://ocw.nctu.edu.tw/upload/classbfs1209122139184046.pdf

http://www.math.mcgill.ca/labute/courses/255w03/L1.pdf

$\endgroup$
1
$\begingroup$

Let $u=x^2$ now integrate $(1+u)^\frac{1}{2}du$,

after integrating sub back in $u=x^2$. thats it!

note: theres no need to find relation between $u=x^2$ as in $\frac{du}{dx}=2x$ shouldnt be substituted in.

$\endgroup$
  • $\begingroup$ can someone verify if my comment is right or wrong :) $\endgroup$ – redrum Oct 25 '12 at 10:25
  • $\begingroup$ It's correct. Some differential calculus tells you that $d(x^2)=2xdx$. We can then substitute $u = x^2$, making $du = 2xdx$, and your integral results. $\endgroup$ – Lord_Farin Oct 25 '12 at 10:35

Your Answer

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