Odd / Even integrals My textbook doesn't really have an explanation for this so could someone explain this too me. 
If f(x) is even, then what can we say about: $$\int_{-2}^{2} f(x)dx$$
If f(x) is odd, then what can we say about $$\int_{-2}^{2} f(x)dx$$ 
I guessed they both are zero? For the first one if its even wouldn't this be the same as $$\int_{a}^{a} f(x)dx = 0$$
Now if its odd f(-x) = -f(x). Would FTOC make this zero as well? 
 A: If $f(x)$ is even then $f(-x) = f(x)$. So $$\int_{-2}^2 f(x) \, \mathrm{d}x = \int_{-2}^0 f(x)\, \mathrm{d}x + \int_0^2 f(x) \, \mathrm{d}x = \int_0^2 f(-x) \, \mathrm{d}x + \int_0^2 f(x) \, \mathrm{d}x$$
But then $f(-x) = f(x)$ so that simplifies to $2\int_0^2 f$. 
Similarly, if $f$ is odd - that is: $f(-x) = -f(x)$ we get $$\int_{-2}^2 f(x) \, \mathrm{d}x = \int_{-2}^0 f(x)\, \mathrm{d}x + \int_0^2 f(x) \, \mathrm{d}x = \int_0^2 f(-x) \, \mathrm{d}x + \int_0^2 f(x) \, \mathrm{d}x = 0$$
A: Not exactly:
$$\begin{cases}\displaystyle\int_{-a}^a f(x)\,\mathrm d\mkern1mu x=2\int_{0}^a f(x)\,\mathrm d\mkern1mu x &\text{if }\;f\;\text{ is even,}\\
\displaystyle\int_{-a}^a f(x)\,\mathrm d\mkern1mu x=0&\text{if }\;f\;\text{ is odd.}\end{cases}$$
To see it, make the substitution $\;t=-x$, $\;\mathrm d\mkern1mu x=-\mathrm d\mkern1mu t$:
$$\int_{-a}^0 f(x)\,\mathrm d\mkern1mu x=-\int_{a}^0 f(-t)\,\mathrm d\mkern1mu t=\int_{0}^a f(-t)\,\mathrm d\mkern1mu t=\begin{cases}\displaystyle\int_{0}^a f(-t)\,\mathrm d\mkern1mu t&(f\;\text{even}),\\\displaystyle-\int_{0}^a f(-t)\,\mathrm d\mkern1mu t&(f\;\text{odd}),\end{cases}$$
then use Chasles relation.
A: Start by splitting the integral into two pieces, the part over negatives values of $x$ and the part over positive values.
$$ \int_{-2}^{2} f(x)\,dx = \int_{-2}^{0} f(x)\,dx + \int_{0}^{2} f(x)\,dx$$
From here you can apply the definition of an even or odd function
A: $$
\int_a^b f(x) dx = \int_a^0 f(x)dx + \int_0^bf(x)dx
$$
if and only if $0 \in (a,b)$ i.e. $0$ is in the interval of your integral. So $a=-2$ and $b=2$ satisfies this easily.
$$
\int_{a}^0 f(x)dx = -\int_{-a}^0 f(-x)dx = \int_0^{-a}f(-x)dx
$$
now since we have the requirement that $0$ is in the interval then we must have $a < 0$ and $b>0$ this implies $-a > 0$ (easy to see) 
putting this together we have
$$
\int_a^b f(x) dx = \int_a^0 f(x)dx + \int_0^bf(x)dx = \int_0^{-a}f(-x)dx + \int_0^bf(x)dx
$$
if we have symmetric bounds i.e. $|a| = |b|$ or $a = - b$ then we have
$$
\int_{-b}^{b} f(x) dx = \int_0^{b}f(-x)dx + \int_0^bf(x)dx = \int_0^b f(-x) + f(x) dx
$$
The final part is what is the parity of a function, the example we have here is odd/even in this sense
$$
f(-x) = -f(x)\;\;\text{odd}\\
f(-x) = f(x)\;\;\text{even}
$$
so we can replace this in the integral.
$$
\int_{-b}^{b} f(x) dx = \int_0^b -f(x) + f(x) dx = \int_0^b 0 dx\;\;\text{odd}\\
\int_{-b}^{b} f(x) dx = \int_0^b f(x) + f(x) dx = \int_0^b 2f(x) dx\;\;\text{even}
$$
A: The intuition for this comes from the pictures (although it could also be proved rigorously). If a function is even then it is symmetrical with respect to the y-axis. Therefore when you integrate it you only need to integrate half of it (greater than zero part or less than zero part) and double your answer.
If the function is odd, it is also symmetric with respect to the y-axis expect this time one side is the negative of the other. This means that when add the integral of the two halves together they will cancel out and you will get zero as your answer.
