Definite integration, even function. confused about proof

[InaudibleTree]

Homework Statement

Let $f$ be integrable on the closed interval $[-a,a]$

If $f$ is an even function, then
$\int^a_{-a}f(x)\,dx$ = $2\int^a_0f(x)\,dx$

Prove this.

Homework Equations

The Attempt at a Solution

The solution is given in the book.

Because $f$ is even, you know that $f(x) = f(-x)$. Using the substitution $u = -x$ produces

$\int^0_{-a}f(x)\,dx =\int^0_af(-u)\,(-du) = -\int^0_af(u)\,du<br /> =\int^a_0f(u)\,du = \int^a_0f(x)\,dx$

Now that part that confuses me is $\int^a_0f(u)\,du = \int^a_0f(x)\,dx$

Wouldnt $u=-x$ mean $du=-dx$ which would produce $\int^a_0f(u)\,du = -\int^a_0f(x)\,dx$

I know my reasoning must fall apart somewhere, since that would mean

If $f$ is an even function, then $\int^a_{-a}f(x)\,dx = 0$.

I just cannot see how my reasoning is wrong.

If it makes any difference this is the remainder of the proof in the book.

$\int^a_{-a}f(x)\,dx =\int^0_{-a}f(x)\,dx + \int^a_0f(x)\,dx<br /> =\int^a_0f(x)\,dx + \int^a_0f(x)\,dx = 2\int^a_0f(x)\,dx$

 

[voko]

You can label the integration variable in a definite integral with any letter, it does not change anything. The definite integral is completely determined by the function under the integral sign and its limits. $\int_a^b f(x)dx = \int_a^b f(u)du = \int_a^b f(\xi)d\xi = \int_a^b f(\mathfrak{S})d\mathfrak{S}$

 

[qbert]

Now that part that confuses me is $\int^a_0f(u)\,du = \int^a_0f(x)\,dx$

Wouldnt $u=-x$ mean $du=-dx$ which would produce $\int^a_0f(u)\,du = -\int^a_0f(x)\,dx$

No because switching $u \leftrightarrow x$ you also have to switch the limits of
integration, which is what you did in the first place but don't do here!

$u=-x$ mean $du=-dx$ which would produce $\int^a_0f(u)\,du = -\int^{-a}_0 f(x)\,dx$

 

[InaudibleTree]

Oh right. I see the mistake I made. Thanks qbert.

But I still don't see how that leads to $\int^0_{-a}f(x)\,dx = \int^a_0f(x)\,dx$

You can label the integration variable in a definite integral with any letter, it does not change anything. The definite integral is completely determined by the function under the integral sign and its limits. $\int_a^b f(x)dx = \int_a^b f(u)du = \int_a^b f(\xi)d\xi = \int_a^b f(\mathfrak{S})d\mathfrak{S}$

I was looing at it as if the author was using a theorem he proved earlier in the text:

If $u=g(x)$ has a continuous derivative on the closed interval $[a,b]$ and $f$ is continuous on the range of $g$, then

$\int^b_af(g(x))g'(x)\,dx = \int^{g(b)}_{g(a)}f(u)\,du$

Im not completley sure but, it would seem the author did use it until he got to the point $\int^a_0f(u)\,du = \int^a_0f(x)\,dx$

Are you saying that when the author got to that point he didnt change the variable, but just relabled it?

 

[voko]

A real change of variables typically changes both the integrand and the limits. This is what the author had till the final step. Then, yes, he simply re-labeled the variable. How can you tell? It did not change the integrand, nor the limits.