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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

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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

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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.

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

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thanks very much @Eivind –  dato datuashvili Aug 3 '11 at 11:59

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