Animated Solution for Mathematics - Definite Integration: If the area (in sq. units) of the region $\{(x,y):y^2\le 4x,x+y\le 1,x\ge 0,y\ge 0\}$ is $a\sqrt{2}+b$, then $a-b$ is equal to :

A Geometric Journey

Welcome, future engineer. Today, we are not just solving an area problem; we are embarking on a journey to understand the delicate interplay between algebra and geometry.

We are tasked with finding the area of a region defined by the inequalities $\{(x,y):y^2\le 4x,x+y\le 1,x\ge 0,y\ge 0\}$. This is a classic JEE Advanced challenge that tests your ability to visualize, analyze, and execute with precision.

Visualizing the Arena

Before we touch a single integral sign, let us paint the picture. We have a parabola $y^2=4x$, which opens to the right.

Since we are restricted to $y\ge 0$, we focus only on the upper branch, $y=2\sqrt{x}$. Then, we have the line $x+y=1$, which cuts through the first quadrant.

The region is bounded by these two curves and the coordinate axes. It is a small, elegant shape nestled in the corner of the first quadrant.

The challenge is that the 'ceiling' of this region changes. For a portion of the $x$-axis, the parabola is the upper boundary, but after a certain point, the line takes over. This is our pivot point.

The Intersection (The Pivot)

To find where the boundary switches, we must find the intersection of $y=2\sqrt{x}$ and $y=1-x$. Equating them gives $2\sqrt{x}=1-x$.

Squaring both sides, we get $4x=(1-x{)}^2$, which simplifies to $x^2-6x+1=0$. Using the quadratic formula, we find $x=3\pm 2\sqrt{2}$.

As we discussed, $3+2\sqrt{2}$ is outside our domain, so our pivot point is $x_0=3-2\sqrt{2}$. This is the moment of truth where the geometry shifts.

The Calculus (Splitting the Burden)

Now, we split the area into two manageable integrals. The total area $A$ is the sum of $A_1$ (the area under the parabola) and $A_2$ (the area under the line):

For $A_1$, we integrate $2x^{1/2}$, which gives $\frac{4}{3}{x}^{3/2}$. Evaluating this at $3-2\sqrt{2}$ looks daunting, but here is where the magic happens.

We recognize that $3-2\sqrt{2}=(\sqrt{2}-1{)}^2$. When we raise this to the power of $3/2$, the square and the half-power cancel out, leaving us with $\frac{4}{3}(\sqrt{2}-1{)}^3$.

Expanding this using the binomial identity $(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3$, we get $5\sqrt{2}-7$. Thus, $A_1=\frac{4}{3}(5\sqrt{2}-7)=\frac{20\sqrt{2}}{3}-\frac{28}{3}$.

The Algebra (The Final Stretch)

For $A_2$, we integrate $(1-x)$ from $3-2\sqrt{2}$ to $1$. The integral is $x-\frac{x^2}{2}$.

Substituting the limits, we get $(1-1/2)-[(3-2\sqrt{2})-\frac{(3-2\sqrt{2}{)}^2}{2}]$. After expanding the square and simplifying the arithmetic, we find $A_2=6-4\sqrt{2}$.

Finally, we combine them: $A=A_1+A_2=(\frac{20\sqrt{2}}{3}-\frac{28}{3})+(6-4\sqrt{2})$.

Grouping the terms, we get $A=\frac{8}{3}\sqrt{2}-\frac{10}{3}$. Comparing this to $a\sqrt{2}+b$, we find $a=8/3$ and $b=-10/3$.

The final value $a-b=8/3-(-10/3)=18/3=6$.

And there you have it! A problem that seemed like a mountain of algebra turned into a beautiful, structured dance of curves and constants.

You have successfully navigated the intersection, the integration, and the algebraic simplification. Keep this mindset—break the problem down, look for the patterns, and trust the process. You are ready for the next challenge.