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When I plug something like this into Mathematica:

$$\int_0^{x^2-1} k y \, dy$$

I get exactly what I would expect:

$$\frac{k}2 (x^2-1)^2 $$

However, when I change my bounds ever so slightly, from $x^2-1$ to $1-x^2$ I would expect this:

$$\frac{k}2 (1-x^2)^2 $$

But I actually end up with the same as before:

$$\frac{k}2 (x^2-1)^2 $$

I'm at a loss as to what I'm missing. I ran these through WolframAlpha as well and got the same results, so I must be missing some basic rule of integration. For reference, here are the Mathematica commands I'm running:

Integrate[k y, {y, 0, (x^2 - 1)}]
Integrate[k y, {y, 0, (1 - x^2)}]
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2 Answers

up vote 3 down vote accepted

Rewrite $(x^2-1)^2$ as $((-1)(1-x^2))^2$ to convince you it's the same.

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Dah! So obvious! Thanks. –  Finster Oct 28 '12 at 23:26
    
No problem! glad to help –  Jean-Sébastien Oct 28 '12 at 23:33
    
Folks seem to forget that the negative of $a-b$ is $b-a$. –  Lubin Oct 29 '12 at 0:42
    
In this case it wasn't so much that I forgot $a-b$ is $b-a$, but that $(x)^2 = (-x)^2$ –  Finster Nov 8 '12 at 18:01
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The two are the same since $(-x)^2 = x^2$.

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