Animated Solution for Mathematics - Definite Integration: Let $f(x)=\{\begin{array}{ll}-2,& -2\le x\le 0\\ x-2,& 0<x\le 2\end{array}$ and $h(x)=f(|x|)+|f(x)|$. Then ${\int }_{-2}^{2}h(x)dx$ is equal to :

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a problem that looks like a standard calculus exercise but is, in reality, a beautiful study of symmetry and piecewise behavior.

We are tasked with evaluating the integral ${\int }_{-2}^{2}h(x)dx$, where $h(x)=f(|x|)+|f(x)|$ and $f(x)$ is a piecewise function defined as:

Visualizing the Chameleon

Our function $f(x)$ is a chameleon. It changes its nature at the origin.

For the domain $x\in [-2,0]$, it is a steady, flat line: $f(x)=-2$. It is unmoving, a constant value.

But as soon as we cross into the positive territory, $x\in (0,2]$, it transforms into a sloping line: $f(x)=x-2$. Imagine drawing this: a horizontal segment from $(-2,-2)$ to $(0,-2)$, and then a line rising from $(0,-2)$ to $(2,0)$.

The Vanishing Act on $[0,2]$

Now, let us tackle $h(x)=f(|x|)+|f(x)|$ on the right side of the axis, where $x\in [0,2]$. Since $x$ is positive, $|x|=x$.

Thus, $h(x)=f(x)+|f(x)|$. We know that for $x\in [0,2]$, $f(x)=x-2$.

Because $x\le 2$, the value of $x-2$ is always non-positive. Therefore, the absolute value $|f(x)|=|x-2|$ must be $-(x-2)$, which is $2-x$.

When we add these together, we get:

The function $h(x)$ effectively vanishes on this interval! It is a flat line on the $x$-axis.

The Linear Growth on $[-2,0]$

Now, let us shift to the left side, where $x\in [-2,0)$. Here, $x$ is negative, so $|x|=-x$.

This means $f(|x|)=f(-x)$. Since $x$ is between $-2$ and $0$, $-x$ is between $0$ and $2$.

We must use the second branch of our original function: $f(-x)=(-x)-2=-x-2$. Meanwhile, for $x\in [-2,0)$, $f(x)=-2$, so $|f(x)|=|-2|=2$.

Combining these, we find:

How elegant! The function $h(x)$ is simply the line $y=-x$ on this interval.

The Final Integration

We have successfully demystified $h(x)$. It is $-x$ from $-2$ to $0$, and $0$ from $0$ to $2$.

The integral is now a simple calculation of area:

The second part is zero. The first part is the area of a right-angled triangle with base $2$ and height $2$.

Using the formula $\frac{1}{2}\times \text{base}\times \text{height}$, we get:

Alternatively, calculating the definite integral:

The final result is 2. We have navigated the piecewise traps and arrived at the solution with clarity and precision.