Question

How do I do this?

If f(x) is an even function, that is, if f(-x)=f(X), show geometrically or otherwise that `int_-a^af(x)dx=`2`int_0^af(x)dx`

Answer 1

See below.

Explanation:

If `f(-x)=f(x),` the graph of the function will be symmetric about the `y-`axis. `Y-`axis symmetry looks like this:

graph{y=-x^2+4 [-10, 10, -5, 5]}

Now, recall that a definite integral on `[a,b]` gives the area under the curve from `[a,b]`.

We're integrating over `[-a, a].` That is, if we split our interval up at `0,` we'll have `[-a, 0], [0, a]`.

Now, if we integrate an even function where `f(-x)=f(x)` over such an interval, the area under `[-a, 0]` will be equal to the area under `[0,a]` due to the function being symmetric about the `y-`axis.

Since these areas are equal, we can say the area under `[-a, a]` is two times the area under `[0, a]`, and thus

`int_-a^af(x)dx=2int_0^af(x)dx`