Definite integral of an odd function is 0 (symmetric interval)

For an odd function, I know that f(x) = - f(x).

I'm trying to show that $\int^{a}_{-a} f(x) dx$ = 0.

I've seen the proof where it splits the integral up into:

$$\int^{a}_{0} f(x) dx + \int^{0}_{-a} f(x) dx$$ However, I still don't understand how to evaluate the second part, which is the 'gold' of the proof.

For example in https://proofwiki.org/wiki/Definite_Integral_of_Odd_Function, why do we define a function from $x \mapsto -x$? (hence leading onto du = - dx and so on). I've been to different sites/videos and they all show me 'how', but not 'why' certain steps are carried out.

Additionally, why can't I evaluate the original integral right away?

$$\int^{a}_{-a} f(x) dx = [F(x)]^{a}_{-a} = F(a) - F(-a) = F(a) - F(a) = 0$$? Is it because we're limited by the need to prove that the integral of that (odd) function is even?

• No, you don't know in a priori that $F$ is an even function. In fact, since you can have $G=F+c$ (where c is a constant), it is even harder to argue that some $F$ is even. (No pun intended) The proofwiki use method of substitution to show that $\int ^{b}_{a} f dx = - \int ^{a}_{b} f dx$. – MonkeyKing Apr 12 '15 at 6:13

Because $f$ is odd, $f(-x) = -f(x)$. Thus, we can write

$$\int_{x=-a}^0 f(x) \, dx = \int_{x=-a}^0 -f(-x) \, dx$$

Let $u = -x, du = -dx$, then

$$\int_{x=-a}^0 -f(-x) \, dx = \int_{u=a}^0 f(u) \, du$$

Reversing the limits of integration inverts the result, so

$$\int_{u=a}^0 f(u) \, du = -\int_{u=0}^a f(u) \, du$$

as desired. (The $u$ is just a dummy variable; it can be replaced by $x$ at this point, despite our originally having assigned it as $u = -x$.)

• Hey i know i'm late here but would you mind explaining this notion of dummy variables, i'm doing this proof in my studies and i notice this idea often but i don't seem to get it. We have changed the limits in terms of u putting u back in terms of x wouldn't that require us to change the limits back?? – Dead_Ling0 Apr 2 at 12:24
• @Dead_Ling0: It's a tricky concept to get, to be sure, but the answer to your last question is no. :-) The idea is that in something like $\int_{u=a}^b f(u)\,du$, the $u$ is just a temporary variable that ranges from $a$ to $b$, and gets plugged into $f$ as an input. Once it does its job, it's discarded. Therefore, it can be renamed whatever we like. Thus, for example, $\int_{u=a}^b f(u)\,du = \int_{x=a}^b f(x)\,dx = \int_{\heartsuit=a}^b f(\heartsuit)\,d\heartsuit$, or whatever (provided $\heartsuit$ is permitted as a variable name). Only the $a$, $b$, and $f$ are "real." – Brian Tung Apr 2 at 16:04
• @Dead_Ling0: If you're familiar with the usual programming languages, you should recognize that for (x = 0; x < 10; x++) { for (y = 0; y < 5; y++) { foo(x, y); } } does exactly the same thing as for (y = 0; y < 10; y++) { for (x = 0; x < 5; x++) { foo(y, x); } }. Same idea. P.S. Shout-out to XKCD 55. – Brian Tung Apr 2 at 16:08