Is this integral correct (even odd functions)

Question

ORIGINAL QUESTION: Suppose that $f(x)$ is an even function and $\int_0^3 f(x)dx=10$. Find $\int_{-3}^0 f(x)dx$.

What about if it is an odd function?

MY ANSWER: If the function is even wouldn't the answer still be $10$. and if the function is odd the answer will be $-10$?

I'm having some troubles with even and odd functions, any help would be greatly appreciated.

Answer 1

Hint : $\int_{-3}^0 f(x) dx = - \int_3^0 f(-x) dx = \int_0^3 f(-x) dx$

Answer 2

Case 1 -- $f$ is even : If $f$ is even, then for all $x$, we have $f(x) = f(-x)$. Then : $$\int_{-c}^0 f(x) dx = \int_c^0 f(-y) (- dy) = \int_0^c f(x) dx$$

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Case 2 -- $f$ is odd : If $f$ is odd, then for all $x$, we have $f(-x) = -f(x)$. Then : $$\int_{-c}^0 f(x) dx = \int_c^0 f(-y) (- dy) = -\int_0^c f(x) dx$$

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In both cases, we have simply made a change of variable from $x$ to $(-x)$. I guess this clarifies your doubt.