Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{-2}^2|1-x^2|dx$ is $\dots \dots \dots$

The goal is to evaluate the definite integral:

The function $f(x)=|1-x^2|$ represents a parabola $1-x^2$ that has been reflected across the $x$-axis wherever it dips below zero. This creates a 'W' shape on the Cartesian plane, crossing the $x$-axis at $x=-1$ and $x=1$.

Before performing the calculation, we observe that $f(-x)=|1-(-x{)}^2|=|1-x^2|=f(x)$. This confirms that the function is even, meaning it is perfectly symmetrical about the $y$-axis.

We can simplify the integral using the property ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$. Thus, our integral becomes:

The modulus function changes behavior at the roots of the internal expression. Since $1-x^2=0$ at $x=1$, we must split the integral at this critical point within the interval $[0,2]$.

For $0\le x\le 1$, the expression $1-x^2$ is non-negative, so $|1-x^2|=1-x^2$. For $1<x\le 2$, the expression is negative, so $|1-x^2|=-(1-x^2)=x^2-1$.

The integral is now expressed as:

We evaluate the two polynomial integrals separately. The antiderivative of $1-x^2$ is $x-\frac{x^3}{3}$, and the antiderivative of $x^2-1$ is $\frac{x^3}{3}-x$.

Evaluating the first part:

Evaluating the second part:

Summing these results and multiplying by the symmetry factor of $2$:

The total area trapped under the curve is 4.