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&#039;(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.