Definite integration, even function. confused about proof

[InaudibleTree]

Homework Statement

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

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

Prove this.

Homework Equations

The Attempt at a Solution

The solution is given in the book.

Because [itex]f[/itex] is even, you know that [itex]f(x) = f(-x)[/itex]. Using the substitution [itex]u = -x[/itex] produces

[itex]\int^0_{-a}f(x)\,dx =\int^0_af(-u)\,(-du) = -\int^0_af(u)\,du
=\int^a_0f(u)\,du = \int^a_0f(x)\,dx[/itex]

Now that part that confuses me is [itex]\int^a_0f(u)\,du = \int^a_0f(x)\,dx[/itex]

Wouldnt [itex]u=-x[/itex] mean [itex]du=-dx[/itex] which would produce [itex]\int^a_0f(u)\,du = -\int^a_0f(x)\,dx[/itex]

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

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

I just cannot see how my reasoning is wrong.

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

[itex]\int^a_{-a}f(x)\,dx =\int^0_{-a}f(x)\,dx + \int^a_0f(x)\,dx
=\int^a_0f(x)\,dx + \int^a_0f(x)\,dx = 2\int^a_0f(x)\,dx[/itex]

 

[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 [itex]\int^a_0f(u)\,du = \int^a_0f(x)\,dx[/itex]

Wouldnt [itex]u=-x[/itex] mean [itex]du=-dx[/itex] which would produce [itex]\int^a_0f(u)\,du = -\int^a_0f(x)\,dx[/itex]

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

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

 

[InaudibleTree]

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

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

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

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

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

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.