Animated Solution for Mathematics - Definite Integration: The area bounded by the curves $y=|x|-1$ and $y=-|x|+1$ is

Welcome, future engineers! Today, we are not just solving a math problem; we are learning to see. In the high-stakes environment of JEE Advanced, the difference between a good rank and a top rank often lies in your ability to visualize the problem before you even touch your pen to paper.

Let’s dive into the beautiful symmetry of the area bounded by $y=|x|-1$ and $y=-|x|+1$.

Imagine a blank coordinate plane. This is our canvas. We are dealing with two functions, both rooted in the absolute value function, $y=|x|$. We know that $y=|x|$ is the quintessential V-shape, with its vertex anchored firmly at the origin $(0,0)$.

Consider the first curve: $y=|x|-1$. That $-1$ outside the modulus is a transformation that pulls the entire V-shape down by exactly one unit. The vertex, once at the origin, now sits at $(0,-1)$.

If we set $y=0$ to find the x-intercepts, we get $|x|=1$, which means $x=1$ and $x=-1$. Our V-shape cuts through the x-axis at these two points.

Now, look at the second curve: $y=-|x|+1$. The negative sign in front of the $|x|$ acts like a mirror, flipping our V-shape upside down. It is now an inverted V, or a peak.

The $+1$ shifts this peak upwards by one unit. So, the vertex is now at $(0,1)$. Again, setting $y=0$ gives us $-|x|+1=0$, or $|x|=1$, leading to $x=\pm 1$.

Do you see it? Both curves pass through $(1,0)$ and $(-1,0)$. They are perfectly aligned, dancing around the x-axis.

When you plot these two curves, you see a shape trapped between them. It is a closed, symmetric quadrilateral. Because the slopes of the lines forming the V-shapes are $1$ and $-1$, the sides of this shape are perpendicular.

This is not just any quadrilateral; it is a rhombus, and specifically, a square tilted by 45 degrees. In geometry, we know that the area of a rhombus is given by the elegant formula:

where $d_1$ and $d_2$ are the lengths of the diagonals. In our case, the diagonals lie perfectly along the x and y axes. The horizontal diagonal, $d_1$, stretches from $x=-1$ to $x=1$, giving it a length of $2$. The vertical diagonal, $d_2$, stretches from $y=-1$ to $y=1$, also giving it a length of $2$.

Now, we simply plug these values into our formula:

And there it is! The area is exactly $2$ square units.

As you prepare for the exam, remember this: patterns are your best friends. For any curves of the form $y=|x|-a$ and $y=a-|x|$, the bounded area will always be $2a^2$.

Here, $a=1$, so $2(1{)}^2=2$. Keep this shortcut in your mental toolkit, but never forget the geometric intuition that derived it. That intuition is what will help you when you face a problem you have never seen before.