Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region $\{(x,y):x\ge 0,x+y\le 3,x^2\le 4y\text{ and }y\le 1+\sqrt{x}\}$ is:

Setting the Stage

Imagine you are standing on the Cartesian plane, looking at a plot of land defined by four distinct fences. Our goal is to find the area of the region enclosed by these fences.

The first constraint, $x\ge 0$, tells us we are strictly in the right half of the plane. The second, $x+y\le 3$, is a straight line, which we can rewrite as $y\le 3-x$. This line acts as a ceiling, sloping downwards from $(0,3)$ to $(3,0)$.

The third constraint, $x^2\le 4y$, is a beautiful, upward-opening parabola, $y\ge \frac{x^2}{4}$, which acts as our floor. Finally, the fourth constraint, $y\le 1+\sqrt{x}$, introduces a gentle, rising curve that starts at $(0,1)$. Our region is the space trapped between these four curves.

To calculate the area, we need to know where these curves meet. The 'roof' of our region is not a single function; it changes.

First, let's find where the root curve $y=1+\sqrt{x}$ meets the line $y=3-x$. Setting them equal, we get $1+\sqrt{x}=3-x$, which simplifies to $\sqrt{x}=2-x$.

Squaring both sides yields $x=4-4x+x^2$, or $x^2-5x+4=0$. Factoring gives $(x-1)(x-4)=0$. Looking at our graph, the intersection happens at $x=1$.

Next, we find where the line $y=3-x$ meets the parabola $y=\frac{x^2}{4}$. Setting $3-x=\frac{x^2}{4}$ leads to $x^2+4x-12=0$, which factors to $(x+6)(x-2)=0$. Since we are in the positive $x$ region, the intersection is at $x=2$.

Because the upper boundary switches from the root curve to the line at $x=1$, we must split our integral into two distinct parts. The lower boundary, the parabola $y=\frac{x^2}{4}$, remains consistent throughout.

Our total area is the sum of the area under the root curve from $0$ to $1$ and the area under the line from $1$ to $2$, minus the area under the parabola from $0$ to $2$. Mathematically, this is:

Let's solve these piece by piece. The first integral evaluates as follows:

The second integral evaluates as follows:

Finally, the area under the parabola is:

Combining these, we get:

The area of our region is exactly $\frac{5}{2}$ square units. You have successfully navigated the boundaries and mastered the calculus of the region!