Evaluate $\int^{2}_{0}f(x)dx$

Question

Let $f(x)=f(-x)$ for $x\in \mathbb{R}$. If $\int^{3}_{-3}f(x)dx=0$ and $\int^{3}_{-2}f(x)dx=5$, then $\int^{2}_{0}f(x)dx=...$

First I tried this:

$\begin{aligned} \int^{3}_{-3}f(x)dx&=2\int^{3}_{0}f(x)dx\\ \int^{3}_{0}f(x)dx&=0\\\\ \int^{3}_{-2}f(x)dx&=\int^{0}_{-2}f(x)dx+\int^{3}_{0}f(x)dx\\ &=\int^{0}_{-2}f(x)dx \end{aligned}$

Let $u=-x$, then $du=-dx$.

$\begin{aligned} -\int^{0}_{2}f(x)dx&=\int^{2}_{0}f(x)dx\\ &=5 \end{aligned}$

Therefore: $\begin{aligned} \int^{2}_{0}f(x)dx&=5 \end{aligned}$

But I didn't feel it's the right method. From the first hypothesis, we know that $f(x)$ is an even function. To satisfy $\int^{3}_{-3}f(x)dx=0$ then $f(x)$ should be $f(x)=0$. But if I assume $f(x)=0$ then it won't satisfy $\int^{3}_{-2}f(x)dx=5$. My questions are:

1. Are there other even function that satisfy $\int^{3}_{-3}f(x)dx=0$ other than $f(x)=0$? As far as I know it's only applied at odd function.
2. If not can I use the first method or there's another method to solve the problem?

Answer 1

$$\int_{-3}^3\cos\frac{\pi x}{3}dx=0$$

Answer 2

What you have done is correct. There are lots of non-zero even functions with $\int_{-3}^{3} f(x)dx=0$. For example we may have $f(x)=\cos(\pi x)$ 0n $(0,3)$.

Answer 3

Here is a very simple function satisfying your condition:

$$f(x) = \begin{cases} 0 & |x|>3 \\ \frac{5}{2} & -2 \leq x \leq 2 \\ -5 & -3 \leq x < -2 \mbox{ and } 2 <x \leq 3 \end{cases}$$