1. Given functions

We have

$$f(x)= \begin{cases} -2, & -2\le x\le 0 \\ x-2, & 0<x\le 2 \end{cases}$$

and

$$h(x)=f(|x|)+|f(x)|.$$

We need to evaluate

$$\int_{-2}^{2} h(x)\,dx.$$

2. Find $f(|x|)$

Since $|x|\in[0,2]$ for $x\in[-2,2]$:

• if $x=0$, then $f(|x|)=f(0)=-2$,
• if $x\ne 0$, then $|x|>0$, so we use the second branch: $$f(|x|)=|x|-2.$$

Thus, for integration purposes,

$$f(|x|)=|x|-2 \quad \text{(except at }x=0\text{, where it is }-2\text{)}.$$

A single point does not affect the integral.

3. Find $|f(x)|$ piecewise

• For $-2\le x\le 0$:

  $$f(x)=-2 \implies |f(x)|=2.$$

• For $0<x\le 2$:

  $$f(x)=x-2.$$

  Since $x\in(0,2]$, we have $x-2\le 0$, so

  $$|f(x)|=|x-2|=2-x.$$

4. Compute $h(x)$ piecewise

For $-2\le x\le 0$

Here $|x|=-x$, so

$$f(|x|)=|x|-2=-x-2.$$

Also,

$$|f(x)|=2.$$

Hence

$$h(x)=(-x-2)+2=-x.$$

For $0<x\le 2$

Here $|x|=x$, so

$$f(|x|)=x-2.$$

Also,

$$|f(x)|=2-x.$$

Hence

$$h(x)=(x-2)+(2-x)=0.$$

So,

$$h(x)= \begin{cases} -x, & -2\le x\le 0,\\ 0, & 0<x\le 2. \end{cases}$$

(The value at $x=0$ is irrelevant for the integral.)

5. Integrate

Therefore,

$$\int_{-2}^{2} h(x)\,dx =\int_{-2}^{0} (-x)\,dx + \int_{0}^{2} 0\,dx.$$

Now,

$$\int_{-2}^{0} (-x)\,dx =\left[-\frac{x^2}{2}\right]_{-2}^{0} =0-\left(-\frac{4}{2}\right)=2.$$

And,

$$\int_0^2 0\,dx=0.$$

Thus,

$$\int_{-2}^{2} h(x)\,dx=2.$$

6. Check options

The correct option is:

$$\boxed{\text{A: }2}$$