Given f is an even function and that

$$\int_{0}^{5} f(x) dx = 12$$

determine

$$\int_{-5}^{5} f(x) dx$$

and

Given f is an odd function and that

$$\int_{0}^{9} f(x) dx = 4$$

determine

$$\int_{-9}^{9} f(x) dx$$

hmmm I simply double for even functions and with odd functions the areas negate each other right? making it 0 ...Is it right to assume that alot of sin and cos graphs will most likely be odd over one complete cycle?
any help would be appreciated ...thanks.

if $f(x)$ is an even function ...

$\displaystyle \int_{-a}^a f(x) \, dx = 2\int_0^a f(x) \, dx$

if $g(x)$ is an odd function ...

$\displaystyle \int_{-a}^a g(x) \, dx = 0$

now the big question ... do you recall how to determine if a function is even, odd, or neither?

Thanks skeeter,
I think even functions have -(fx) = (fx)
odd -(fx) =(-fx)
and neither doesn't suit those rules ...to do with powers and constants ..a good recap though got me thinking about it...thanks again ...makes things more clearer

If $(\forall x)[f(-x)=f(x)]$ then $f$ is an even function.
If $(\forall x)[f(-x)=-f(x)]$ then $f$ is an odd function.