Animated Solution for Mathematics - Definite Integration: The area (in square units) bounded by the curves $y=\sqrt{x},2y-x+3=0$, x-axis, and lying in the first quadrant is:

Welcome, fellow traveler, to the beautiful world of integral calculus. Today, we are not just solving a problem; we are exploring a geometric landscape.

Imagine standing in the first quadrant of the Cartesian plane. We have two paths defined by our equations: the parabola $y=\sqrt{x}$ (or, more conveniently for us, $x=y^2$) and the straight line $2y-x+3=0$.

Our goal is to find the area trapped between these two curves and the $x$-axis. This is a classic JEE Advanced challenge—it tests not just your ability to integrate, but your ability to strategize.

Before we can calculate, we must understand the boundaries of our territory. Where do these two curves meet?

To find this, we set up a system of equations. We know $x=y^2$ from the parabola. Substituting this into our line equation, $2y-x+3=0$, we get:

Rearranging this, we arrive at the quadratic equation:

Factoring this, we find $(y-3)(y+1)=0$. This gives us two potential intersection points: $y=3$ and $y=-1$.

Since we are restricted to the first quadrant, where $y$ must be positive, we gracefully reject $y=-1$ and accept $y=3$. Plugging this back into $x=y^2$, we find our intersection point at $(9,3)$.

Now, here is where the elite student separates themselves from the crowd. We could integrate with respect to $x$.

But look at the region! If we integrate along the $x$-axis, the lower boundary changes at $x=3$. We would have to calculate two separate integrals: one for the area under the parabola and one for the area under the line.

That is a recipe for a calculation error. Instead, let's integrate with respect to $y$.

By using horizontal strips of thickness $dy$, we see that the right boundary is always the line $x=2y+3$ and the left boundary is always the parabola $x=y^2$. This allows us to calculate the entire area with a single, elegant integral.

With our strategy locked in, the integral becomes a joy to solve. The area is the integral from $y=0$ to $y=3$ of the right boundary minus the left boundary:

Integrating term by term, we get:

Evaluating at the upper limit $y=3$, we get:

Evaluating at the lower limit $y=0$ gives us $0$. Thus, the total area is exactly $9$ square units.

It is a clean, satisfying result. Remember, in JEE Advanced, the most complex-looking problems often yield to the simplest, most elegant strategies. Keep visualizing, keep strategizing, and keep falling in love with the math!