Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For example we want to integrate following integral $$\int_0^4 2x\cdot((9+x^2)^{1/2})\,dx$$ I have read that if we denote $u$ as $u=9+x^2$ then $du=2x*dx$ everything is clear here but then there was used such kind of method: the author simply put $x=0$ and $x=4$ into the original $u$ definiton got $u=9$ and $u=25$ and finally wrote integral as $$\int_{u=9}^{u=25} u^{1/2}\,du.$$ My question is: is it correct? Or if I will use the same method to solve similar integrals next time, will I be wrong or not? thanks

share|cite|improve this question
There may be a little fiddling to do if your change of variables is not one-to-one on the interval of integration. But (check) in this case $u=9+x^2$ is one-to-one where $x$ ranges over $[0,4]$. – GEdgar Aug 3 '11 at 13:10
up vote 1 down vote accepted

It is correct, because you have the following:

$$\int_{a}^{b} f(\varphi(t)) \cdot \varphi'(t)\,\mathrm{d}t = \int_{\varphi(a)}^{\varphi(b)} f(x)\,\mathrm{d}x$$

Hope this helps.

share|cite|improve this answer
thanks very much @Daniel – dato datuashvili Aug 3 '11 at 10:26

The author is correct. Let me try to explain why by using your example. $$\int_{x=0}^{x=4} 2x(9+x^2)^{\frac{1}{2}} \text{d}x$$ If we let $u=9+x^2$ we get, as you said, $\text du = 2x \;\text dx$. So $$\int_{x=0}^{x=4} 2x(9+x^2)^{\frac{1}{2}} \text{d}x = \int_{x=0}^{x=4} u^{\frac{1}{2}} \;\text du.$$

$f(u)= u^{\frac{1}{2}}$ can easily be plotted in a coordinate system with a horizontal $u$-axis and a vertical $y$-axis. If you view the integral as the area under the curve, what does it mean to take the integral from $x=0$ to $x=4$? Remember that $x=0$ is not the same as $u=0$. So we translate from $x$ to $u$ like this: $$ x= 0 \Rightarrow u = 9+0^2 = 0$$ $$ x= 4 \Rightarrow u = 9+4^2 = 25$$

And we get $$\int_{x=0}^{x=4} u^{\frac{1}{2}} \;\text du =\int_{u=9}^{u=25} u^{\frac{1}{2}} \;\text du.$$

share|cite|improve this answer
thanks very much @Eivind – dato datuashvili Aug 3 '11 at 11:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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