1
$\begingroup$

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)}] 

2 Answers 2

3

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

  • 0
    Dah! So obvious! Thanks.2012-10-28
  • 0
    No problem! glad to help2012-10-28
  • 0
    Folks seem to forget that the negative of $a-b$ is $b-a$.2012-10-29
  • 0
    In this case it wasn't so much that I forgot $a-b$ is $b-a$, but that $(x)^2 = (-x)^2$2012-11-08
4

The two are the same since $(-x)^2 = x^2$.