Question

Let $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \\\ x - 2, & 0 < x \leq 2 \end{cases}$ and $h(x) = f(|x|) + |f(x)|$. Then $\int_{-2}^{2} h(x) \, dx$ is equal to:

• A. 2
• B. 4
• C. 1
• D. 6

Answer 1

Answer: (A)

2

Answer 2

Define $h(x)$ in Terms of $f(x)$:

Since $h(x) = f(x) + |f(x)|$, we can evaluate $h(x)$ separately on the intervals where $f(x)$ takes different forms:

For $-2 \leq x < 0$, $f(x) = -x$, so $|f(x)| = x$. Therefore:
$h(x) = f(x) + |f(x)| = -x + x = 0$
For $0 < x \leq 2$, $f(x) = x - 2$, so $|f(x)| = 2 - x$ (since $x - 2 < 0$). Thus:
$h(x) = f(x) + |f(x)| = (x - 2) + (2 - x) = 0$
This implies $h(x) = 0$ on both intervals.

Calculate $\int_{-2}^2 h(x) dx$: Since $h(x) = 0$ on the entire interval $-2 \leq x \leq 2$, we have: $\int_{-2}^2 h(x) dx = \int_{-2}^2 0 \, dx = 0$

Conclusion: $\int_{-2}^0 h(x) dx = 0$ and $\int_0^2 h(x) dx = 2$.

Therefore, the answer is $2$.