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$$\int_{-1}^1 x^2 ~dx = \frac{2}{3}.$$

Now let us substitute $u=x^2$. Then $du=2x ~dx$ then the definite integration becomes from $1$ to $1$, i.e. $0$. I know I need to partition this from $-1$ to $0$ and $0$ to $1$. But my question is when we substitute anything we do not concern ourselves with the fact that what we are substituting is even or odd. Then why do we need to partition this?

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Why that particular substitution? It isn't going to make things very nice.... –  Cameron Buie Aug 13 '12 at 18:33
    
I know but in that case i am getting a different answer which is my problem –  SN77 Aug 13 '12 at 18:41
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up vote 6 down vote accepted

Carry out the substitution. As you point out, we get $du=2x\,dx$. But we need to express everything in terms of $u$, in particular $dx$. We cannot say $2x=2\sqrt{u}$, for that is incorrect when $x$ is negative. So the integral will have to be broken up at $x=0$.

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Roughly speaking the substitution needs to be invertible. In this case, if you do not want to partition the integral, you need to use a different substitution which preserves invertibility; in this case $u=x^2$ for $x\geq 0$, and $u=-x^2$ for $u<0$. So, in effect, you are still partitioning, you have just moved the bump under the carpet somewhere else.

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