Key Concepts and Formulas

1. Piecewise Functions: Understanding how to define and work with functions that have different definitions over different intervals of their domain.
2. Modulus Function Properties: The behavior of $|x|$ and $|f(x)|$, specifically $|a|=a$ if $a \ge 0$ and $|a|=-a$ if $a < 0$.
3. Definite Integration of Piecewise Functions: The integral of a piecewise function over an interval is the sum of the integrals over each sub-interval where the function's definition is constant. $\int_a^b g(x) \mathrm{d}x = \sum_{i=1}^n \int_{a_i}^{a_{i+1}} g_i(x) \mathrm{d}x$ where $a=a_1 < a_2 < \dots < a_{n+1}=b$ and $g(x)=g_i(x)$ on $[a_i, a_{i+1}]$.

Step-by-Step Solution

We are asked to evaluate the definite integral $\int_{-2}^2 \mathrm{~h}(x) \mathrm{d} x$, where $\mathrm{h}(x)=f(|x|)+|f(x)|$. To do this, we first need to determine the explicit form of $h(x)$ over the interval $[-2, 2]$.

Step 1: Analyze the function $f(x)$. The given function $f(x)$ is defined as: $f(x)=\left\{\begin{array}{lr} -2, & -2 \leq x \leq 0 \\ x-2, & 0< x \leq 2 \end{array}\right.$

Step 2: Determine $f(|x|)$ for $x \in [-2, 2]$. We need to substitute $|x|$ into the definition of $f(x)$. For $x \in [-2, 2]$, the value of $|x|$ ranges from $0$ to $2$. Let $t = |x|$.

• If $x \in [-2, 0)$, then $|x| = -x$. Since $-2 \le x < 0$, we have $0 < -x \le 2$. Thus, $t \in (0, 2]$. In this range, $f(t) = t-2$. So, $f(|x|) = |x|-2 = (-x)-2 = -x-2$.
• If $x = 0$, then $|x| = 0$. According to the definition of $f(x)$, $f(0) = -2$. So, $f(|x|) = -2$.
• If $x \in (0, 2]$, then $|x| = x$. Since $0 < x \le 2$, we have $t \in (0, 2]$. In this range, $f(t) = t-2$. So, $f(|x|) = |x|-2 = x-2$.

Combining these, we get $f(|x|)$ for $x \in [-2, 2]$: $f(|x|)=\left\{\begin{array}{lr} -x-2, & -2 \leq x < 0 \\ -2, & x=0 \\ x-2, & 0< x \leq 2 \end{array}\right.$ We can simplify this by noting that when $x=0$, $-x-2 = -0-2=-2$, and $x-2 = 0-2=-2$. So, we can combine the intervals: $f(|x|)=\left\{\begin{array}{lr} -x-2, & -2 \leq x \leq 0 \\ x-2, & 0< x \leq 2 \end{array}\right.$

Step 3: Determine $|f(x)|$ for $x \in [-2, 2]$. We need to consider the sign of $f(x)$ in its defined intervals.

• For $-2 \leq x \leq 0$, $f(x) = -2$. Since $-2$ is negative, $|f(x)| = |-2| = 2$.
• For $0 < x \leq 2$, $f(x) = x-2$.
  • If $0 < x < 2$, then $x-2$ is negative. So, $|f(x)| = |x-2| = -(x-2) = 2-x$.
  • If $x = 2$, then $f(2) = 2-2 = 0$. So, $|f(2)| = 0$. Note that $2-x$ is $0$ when $x=2$, so we can write: $|f(x)|=\left\{\begin{array}{lr} 2, & -2 \leq x \leq 0 \\ 2-x, & 0< x < 2 \\ 0, & x=2 \end{array}\right.$ For integration purposes, the value at a single point does not affect the integral, so we can consider $|f(x)| = 2-x$ for $0 < x \le 2$. Thus, for $x \in [-2, 2]$: $|f(x)|=\left\{\begin{array}{lr} 2, & -2 \leq x \leq 0 \\ 2-x, & 0< x \leq 2 \end{array}\right.$

Step 4: Determine $h(x) = f(|x|) + |f(x)|$ for $x \in [-2, 2]$. We need to combine the definitions of $f(|x|)$ and $|f(x)|$ over the interval $[-2, 2]$. The critical point for splitting the interval is $x=0$.

• Interval 1: $-2 \leq x \leq 0$. In this interval, $f(|x|) = -x-2$ and $|f(x)| = 2$. So, $h(x) = (-x-2) + 2 = -x$.

• Interval 2: $0 < x \leq 2$. In this interval, $f(|x|) = x-2$ and $|f(x)| = 2-x$. So, $h(x) = (x-2) + (2-x) = 0$.

Therefore, the function $h(x)$ over the interval $[-2, 2]$ is: $h(x)=\left\{\begin{array}{lr} -x, & -2 \leq x \leq 0 \\ 0, & 0< x \leq 2 \end{array}\right.$

Step 5: Evaluate the definite integral $\int_{-2}^2 \mathrm{~h}(x) \mathrm{d} x$. We split the integral based on the piecewise definition of $h(x)$: $\int_{-2}^2 \mathrm{~h}(x) \mathrm{d} x = \int_{-2}^0 \mathrm{~h}(x) \mathrm{d} x + \int_{0}^2 \mathrm{~h}(x) \mathrm{d} x$ Substitute the respective definitions of $h(x)$: $= \int_{-2}^0 (-x) \mathrm{d} x + \int_{0}^2 (0) \mathrm{d} x$ Now, we evaluate each integral: $\int_{-2}^0 (-x) \mathrm{d} x = \left[-\frac{x^2}{2}\right]_{-2}^0 = \left(-\frac{0^2}{2}\right) - \left(-\frac{(-2)^2}{2}\right) = 0 - \left(-\frac{4}{2}\right) = 0 - (-2) = 2$ $\int_{0}^2 (0) \mathrm{d} x = 0$ Summing the results: $\int_{-2}^2 \mathrm{~h}(x) \mathrm{d} x = 2 + 0 = 2$

Common Mistakes & Tips

• Careful handling of $|x|$: When substituting $|x|$, remember that $|x|$ is always non-negative. This means you need to consider how the original intervals of $f(x)$ map to the intervals of $|x|$.
• Sign analysis of $f(x)$: Properly determine when $f(x)$ is positive, negative, or zero to correctly evaluate $|f(x)|$.
• Combining piecewise functions: Ensure that when defining $h(x)$, you correctly add the corresponding pieces of $f(|x|)$ and $|f(x)|$ for each interval.

Summary

To solve this problem, we first broke down the definition of $h(x) = f(|x|) + |f(x)|$ by analyzing $f(|x|)$ and $|f(x)|$ separately over the interval $[-2, 2]$. This resulted in a piecewise definition for $h(x)$. We then used the property of definite integrals for piecewise functions to split the integral into sub-integrals over the relevant intervals and evaluated each part. The sum of these integrals yielded the final answer.

The final answer is $\boxed{2}$.