Animated Solution for Mathematics - Definite Integration: The area enclosed between the curves $y=x|x|$ and $y=x-|x|$ is :

Imagine you are standing on a coordinate plane, looking at two curves that seem simple at first glance: $y=x|x|$ and $y=x-|x|$. The modulus function, $|x|$, is a master of disguise.

It behaves differently depending on whether you are in the positive realm ($x\ge 0$) or the negative realm ($x<0$). To find the area trapped between these two, we must first unmask them.

Unmasking the Curves

Let us break down the first curve, $y=x|x|$. When $x\ge 0$, $|x|$ is just $x$, so our curve becomes $y=x^2$, a classic upward-opening parabola.

When we step into the negative territory where $x<0$, $|x|$ transforms into $-x$. Suddenly, our curve becomes $y=x(-x)=-x^2$, a downward-opening parabola. We have two distinct branches meeting at the origin.

Now, look at the second curve, $y=x-|x|$. In the positive realm ($x\ge 0$), $|x|$ is $x$, so $y=x-x=0$. This is just the positive $x$-axis.

In the negative realm ($x<0$), $|x|$ is $-x$, so $y=x-(-x)=2x$. This is a straight line with a slope of $2$. We have successfully deconstructed our curves into piecewise components.

Finding the Meeting Point

To find the area, we need to know where these curves meet. On the positive side ($x\ge 0$), we equate $x^2=0$, which gives us $x=0$. They meet at the origin $(0,0)$.

On the negative side ($x<0$), we equate the two negative branches: $-x^2=2x$. Rearranging this gives us $x^2+2x=0$, or $x(x+2)=0$.

Since we are in the negative region, we ignore $x=0$ and focus on $x=-2$. This is our second intersection point. The curves are trapped between $x=-2$ and $x=0$.

The Calculus of the Trap

Now, we visualize the region. Between $x=-2$ and $x=0$, the parabola $y=-x^2$ sits above the line $y=2x$. This provides our upper and lower boundaries.

The area $A$ is the integral of the upper curve minus the lower curve:

We integrate term by term: the integral of $-x^2$ is $-\frac{x^3}{3}$, and the integral of $-2x$ is $-x^2$. We evaluate this from $-2$ to $0$:

Plugging in the upper limit $0$ gives us $0$. Plugging in the lower limit $-2$ gives us:

Subtracting this from $0$ gives us:

The Beauty of the Result

And there it is: $\frac{4}{3}$. The negative signs cancel out, the arithmetic resolves, and we are left with a clean, positive area.

This problem is about the discipline of breaking down complex functions into manageable pieces. Every time you see a modulus, remember: it is not a wall, but a gateway to a piecewise reality.