Animated Solution for Mathematics - Definite Integration: If the area of the region $\{(x,y):1+x^2\le y\le \min \{x+7,11-3x\}\}$ is $A$, then $3A$ is equal to

My dear student, welcome to a beautiful problem. Often in JEE Advanced, we are presented with functions that seem intimidating at first glance, like the $\min$ function.

But I want you to stop and visualize this. Imagine you are standing on a floor shaped like a parabola, $y=1+x^2$. Above you, there is a roof formed by two intersecting lines, $y=x+7$ and $y=11-3x$.

Your task is to find the area of the space between this floor and this tent.

Before we can calculate any area, we must understand the structure of our roof. The $\min$ function tells us that at any point $x$, the roof is the lower of the two lines.

To find where the roof changes its slope, we must find where these two lines meet:

Solving this, we get $4x=4$, which means $x=1$. This is the peak of our tent! At $x=1$, the roof transitions from the first line to the second.

Now, we need to know where our region starts and ends. The region is bounded below by the parabola $y=1+x^2$. We need to find where this parabola meets our roof.

On the left side, we equate the parabola with the first line:

Factoring this quadratic, we get $(x-3)(x+2)=0$. Looking at our graph, the relevant intersection is at $x=-2$.

On the right side, we equate the parabola with the second line:

This factors to $(x+5)(x-2)=0$. The relevant intersection here is at $x=2$. So, our region of interest spans from $x=-2$ to $x=2$.

Now for the heavy lifting. Because the roof changes at $x=1$, we must split our integral into two parts. The total area $A$ is the sum of two integrals:

Let's simplify the integrands first. The first part becomes ${\int }_{-2}^1(x+6-x^2)dx$, and the second part becomes ${\int }_1^2(10-3x-x^2)dx$.

Integrating the first part, we get:

For the second part, we integrate to get:

Adding these together, we find:

We have found the area $A=\frac{50}{3}$. The question asks for $3A$.

Multiplying our result by $3$, we get:

And there it is! A clean, elegant integer. You have mastered the geometry, and the calculus followed suit. Keep this clarity of thought, and no JEE problem will ever be too difficult for you.