Animated Solution for Mathematics - Definite Integration: Sketch the region bounded by the curves $y=x^2$ and $y=\frac{2}{1+x^2}$. Find the area.

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving an integral; we are witnessing a beautiful interaction between two distinct mathematical entities: the ever-growing parabola $y=x^2$ and the elegant, decaying bell curve $y=\frac{2}{1+x^2}$.

Imagine standing on a coordinate plane. To your left and right, the parabola climbs toward infinity, while the bell curve sits proudly at the origin, reaching a height of $2$ before gracefully tapering off toward the x-axis. Our mission is to find the 'pocket' of space trapped between them.

Before we can measure the space, we must define the boundaries. Where do these two paths cross? We set them equal:

This looks like a daunting quartic equation, but let us breathe. If we multiply both sides by $(1+x^2)$, we get $x^2(1+x^2)=2$, which expands to $x^4+x^2-2=0$.

By treating $x^2$ as a single variable, say $u$, we see the quadratic structure: $u^2+u-2=0$. Factoring this gives $(u+2)(u-1)=0$.

Since $x^2$ cannot be negative, we discard $u=-2$ and keep $u=1$. Thus, our intersection points are $x=1$ and $x=-1$. We have found our gates!

Now, we set up our integral. The area $A$ is the integral of the upper curve minus the lower curve:

Here is where we use our 'JEE intuition.' Both functions are even, meaning they are mirror images across the y-axis. Why calculate the whole thing when we can calculate the right half and double it?

It is cleaner, faster, and less prone to those pesky sign errors. So, we transform our integral into:

Now, we perform the integration term by term. The integral of $\frac{2}{1+x^2}$ is a classic result: $2{\tan }^{-1}(x)$. The integral of $x^2$ is the simple power rule: $\frac{x^3}{3}$.

Putting it all together, we have:

This is the moment of truth. We substitute our limits. At the upper limit $x=1$, we know ${\tan }^{-1}(1)=\frac{\pi }{4}$. At the lower limit $x=0$, both terms vanish into nothingness.

Substituting these values, we get:

Distributing the $2$, we arrive at our final, elegant answer:

Look at that result. It is not just a number; it is a testament to the harmony between geometry and calculus. You started with two complex curves and ended with a precise, finite value.