How prove an equality We have $f(x)=\sqrt{x^2+10}$ and we need to show: $\int_{-10}^{10} f(x) \, dx = 2\int_0^{10} f(x) \, dx$
First method I find the primitive for this function and after I put the value and I obtain this equality.But how we can easily prove this equality?

We have, for all $a\in \mathbb{R}$ (for you $a=10$)
$$\int_{-a}^a f(x) \text{d} x = \int_0^a f(x) \text{d} x + \int_{-a}^0 f(x) \text{d} x .$$
But thanks to a change of variable $u =-x$ 
$$ \int_{-a}^0 f(x) \text{d} x = \int_a^0 -f(-u)\text{d} u = \int_0^a f(-u) \text{d} u.$$
We finally obtain the result since $f$ is even which means that for all $x\in \mathbb{R}$, $f(x)=f(-x)$.
