Definite integral of an odd function If f(x) is odd and $\int_{-2}^{7} f(x) dx$ = 8, then $\int_{2}^{7} f(x) dx$ =__
I have no clue where to even start. I looked up even and odd functions because I forgot what those were, and I figured out at least that for the odd function, the (-2,0) part of the first integral should cancel out some of the area and the area of the second integral should be greater than 8. Can someone give me a hint as to what I can begin with?
 A: Hint: $\int_{-2}^7 f(x)dx=  \int_{-2}^2 f(x)dx + \int_{2}^7 f(x)dx$
A: Based on the given information, you don't know whether 
$\int_{-2}^0 f(x)\,dx$ is positive or negative, and whether "the rest" 
of the integral would be less than or greater than
$\int_{-2}^7 f(x)\,dx$.
Instead, consider $\int_{-2}^2 f(x)\,dx$. Is it positive, negative, or zero?
A: Rewrite the integral as:
$$\int_{-2}^7 f(x)\ \text{d}x = \int_{-2}^2 f(x)\ \text{d}x + \int_2^7 f(x)\ \text{d}x$$
You have an odd function, and a property of odd function is this:
$$\int_{-a}^a f(x)\ \text{d}x = 0$$
This means that 
$$\int_{-2}^2 f(x) \ \text{d}x = 0$$
And you remain with
$$\int_{-2}^7 f(x)\ \text{d}x = \int_2^7 f(x)\ \text{d}x$$
But since you know that the first term on the left side is equal to $8$, then... also the right side is $8$ because of the equality we got.
A: You are close.  You have figured out that in the first integral, the area from $-2$ to $0$ cancels out the area from $2$ to $7$.  So what does that say about the 
integral from $2$ to $7$, compared to the fisrt integral from $-2$ to $7$?
