Verification:$f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and even, $\forall a\in\mathbb{R}_+^*, \int_{-a}^{a} f(t) \,dt=2\int_{0}^{a} f(t) \,dt$ I'd like a proof verification. I tend to forget/miss details. I hope it is right, if not I'd love to here what went wrong.
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous even function. Show that $\forall a\in\mathbb{R}_+^*, \int_{-a}^{a} f(t) \,dt=2\int_{0}^{a} f(t) \,dt$.
Proof:
Because $f$ is even and continuous on $\mathbb{R}$, $\forall x\in\mathbb{R}, f(-x)=f(x)$.
By using Chasles's relation, we have that $\int_{-a}^{a} f(t) \,dt=\int_{-a}^{0} f(t) \,dt+\int_{0}^{a} f(t) \,dt=-\int_{0}^{-a} f(t) \,dt+\int_{0}^{a} f(t) \,dt$. As $f$ is continuous on $\mathbb{R}$, there exists an antiderivative $F(x)$ (such that $F'(x)=f(x),\forall x\in\mathbb{R}$). By using the fundamental theorem, $-\int_{0}^{-a} f(t) \,dt+\int_{0}^{a} f(t) \,dt=(-F(-a)-(-F(0)))+F(a)-F(0)$. [1]
Now, we will show that $\int_{0}^{x} f(t) \,dt$ is odd if $f$ is even. Because $f$ is even,$\int_{0}^{x} f(t) \,dt=\int_{0}^{x} f(-t) \,dt$. let $u=-t$, we have that $\int_{0}^{x} f(-t) \,dt=-\int_{0}^{x} f(u) \,du\implies\int_{0}^{x} f(-t) \,dt=-\int_{0}^{x} f(-t) \,dt\implies$the antiderivative is odd.
Because the antiderivative is odd, [1] becomes $F(a)-F(-0)+F(a)-F(0)=2(F(a)-F(0))=2\int_0^af(t)dt$.
 A: There's a much simpler way you can go about this after rewriting as $-\int^{-a}_0 f(t) dt + \int_0^a f(t) dt.$
For the second integral, simply consider the substitution $u = -t, du = -dt.$ Note that $u(-a) = a, u(0) = 0,$ so we now have $-\int_0^a f(-u) (-du) = \int_0^a f(-u) du.$ Now by the evenness of $f, f(-u) = f(u)$ for all $u,$ so this is simply $\int_0^a f(u) du.$
So, $\int_{-a}^a f(t) dt = \int_0^a f(t) dt + \int_0^a f(u) du = 2\int_0^a f(t) dt.$
A: It is correct!
By the way, it feels nice to see the LaTeX formatting. Nice job (and patience) there!

Although you can simplify the whole thing by just using the substitution $u=-t$ for the first part right away after breaking down the integral so that it becomes
$$\int_0^a f(-u)\,\text du+\int_0^af(t)\,\text dt$$
Since $f$ is even, this is the same as
$$\int_0^a f(u)\,\text du+\int_0^af(t)\,\text dt$$
Since the variable in definite integrals is a dummy variable, so the variable name doesn't matter. The above expression is the same as
$$\int_0^af(t)\,\text dt+\int_0^af(t)\,\text dt$$
$$=2\int_0^af(t)\,\text dt$$
as desired.

Hope this helps. Ask anything if not clear :)
