Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{-2}^3|1-x^2|dx$ is

A Journey Through Integration

Welcome, fellow traveler in the world of calculus! Today, we are going to tackle a problem that might look like a simple integral at first glance, but it hides a beautiful geometric reality.

We are tasked with evaluating the definite integral:

At first, the absolute value symbol $|\cdot |$ might seem like a barrier, but think of it as a gatekeeper. It tells us that the function is always non-negative, effectively 'flipping' any negative parts of the graph above the $x$-axis. To solve this, we must first understand the behavior of the expression inside: $f(x)=1-x^2$.

The Gatekeepers (Critical Points)

Before we dive into the integration, we must find where the function changes its nature. The expression $1-x^2$ is a downward-opening parabola.

It crosses the $x$-axis when $1-x^2=0$, which gives us $x=1$ and $x=-1$. These are our critical points.

Because these points fall within our range $[-2,3]$, we cannot simply integrate the function as one piece. We must respect the boundaries and split our journey into three distinct segments: $[-2,-1]$, $[-1,1]$, and $[1,3]$.

The Piecewise Transformation

Now, let's define our function in each of these segments. In the interval $[-2,-1]$, if you test a value like $x=-1.5$, you will find that $1-x^2$ is negative. Thus, the modulus forces us to flip the sign: $|1-x^2|=x^2-1$.

In the middle interval, $[-1,1]$, the expression $1-x^2$ is positive, so the modulus does nothing: $|1-x^2|=1-x^2$.

Finally, in the interval $[1,3]$, the expression becomes negative again, so we flip it back: $|1-x^2|=x^2-1$. We have successfully transformed our single, intimidating integral into a sum of three manageable pieces:

The Integration Journey

Now, we apply the power rule, $\int x^{n}dx=\frac{x^{n+1}}{n+1}$.

For the first part, $I_1={\int }_{-2}^{-1}(x^2-1)dx$, the antiderivative is $\frac{x^3}{3}-x$. Evaluating this from $-2$ to $-1$:

For the second part, $I_2={\int }_{-1}^1(1-x^2)dx$, the antiderivative is $x-\frac{x^3}{3}$. Evaluating from $-1$ to $1$:

Finally, for the third part, $I_3={\int }_1^3(x^2-1)dx$, we use the same antiderivative as the first part: $\frac{x^3}{3}-x$. Evaluating from $1$ to $3$:

We have reached the end of our journey. All that remains is to sum our three results:

Adding these fractions is straightforward since the denominators are identical:

You have just navigated the complexities of the modulus function and emerged victorious! The final answer is $\frac{28}{3}$.