Animated Solution for Mathematics - Definite Integration: The area enclosed between the curves $y^2=x$ and $y=|x|$ is

Unveiling the Area

My dear student, welcome to a beautiful intersection of geometry and calculus. Today, we are not just solving a problem; we are witnessing a dance between two fundamental shapes: the parabola and the modulus function.

Often, the hardest part of a JEE problem is not the integration itself, but the courage to visualize what is happening on the coordinate plane. Let us peel back the layers together.

Visualizing the Geometry

We are given two curves: $y^2=x$ and $y=|x|$. The first, $y^2=x$, is a classic parabola opening to the right. It is the path of a projectile, the shape of a satellite dish, a curve of infinite grace.

The second, $y=|x|$, is the absolute value function—a sharp, V-shaped graph that reflects perfectly across the $y$-axis.

But here is the crucial insight: look at the parabola's equation, $y^2=x$. Since the square of any real number is non-negative, $x$ must be greater than or equal to zero.

This immediately restricts our entire universe to the right side of the $y$-axis. In this region, the modulus function $y=|x|$ simplifies beautifully to just $y=x$.

So, we are essentially looking for the area trapped between the upper branch of the parabola, $y=\sqrt{x}$, and the straight line $y=x$.

The Meeting Point

Before we can calculate the area, we must know where these two curves meet. They are like two travelers crossing paths in the night. To find these points, we set the functions equal to each other:

Now, do not be tempted to divide by $x$! If you do, you will lose the origin, and your entire calculation will collapse. Instead, square both sides to get $x=x^2$, and then rearrange it into a quadratic equation:

Factoring this gives us $x(x-1)=0$. Our intersection points are $x=0$ and $x=1$. These are the boundaries of our world, the limits of our integration.

The Calculus of Area

Now, we enter the realm of integration. The area $A$ enclosed between two curves is the integral of the 'upper' curve minus the 'lower' curve. As we discovered with our test point, in the interval $[0,1]$, the parabola $y=\sqrt{x}$ sits above the line $y=x$.

Thus, our integral is:

Let us break this down. We use the power rule, $\int x^{n}dx=\frac{x^{n+1}}{n+1}$.

For the first term, $\sqrt{x}$ is just $x^{1/2}$. Its integral is $\frac{x^{3/2}}{3/2}$, which simplifies to $\frac{2}{3}{x}^{3/2}$. For the second term, $x$, the integral is simply $\frac{x^2}{2}$.

Putting it all together, our antiderivative is:

Now, we apply the Fundamental Theorem of Calculus. We substitute the upper limit, $x=1$:

Subtracting the lower limit, $x=0$, gives us zero. So, we are left with the simple arithmetic of fractions. Finding a common denominator of $6$, we get:

And there it is! The area enclosed is exactly $\frac{1}{6}$ square units.

It is a small, elegant number, but it represents the perfect harmony of the two curves. Remember, my student: when you face these problems, do not rush to the formulas. Visualize the curves, respect the domain, and let the calculus flow naturally. You have got this!