Animated Solution for Mathematics - Definite Integration: Let the area of the region $\{(x,y):0\le x\le 3,0\le y\le \min \{x^2+2,2x+2\}\}$ be $A$. Then $12A$ is equal to

Welcome, fellow explorers of mathematics! Today, we are going to unravel a beautiful problem that tests not just your calculus skills, but your ability to visualize the behavior of functions.

We are tasked with finding the area $A$ of a region bounded by $0\le x\le 3$ and $0\le y\le \min \{x^2+2,2x+2\}$, and then calculating $12A$.

The heart of this problem lies in the operator $\min \{x^2+2,2x+2\}$. This is not a single function; it is a composite boundary. The 'minimum' operator tells you to always pick the lower of the two curves, meaning the upper boundary of our region changes its identity at the point where the two curves intersect.

Before we can integrate, we must find where this switch happens. We set the two functions equal to each other:

The constant term $2$ cancels out on both sides, leaving us with the elegant equation $x^2-2x=0$. Factoring this, we get $x(x-2)=0$.

This tells us the curves intersect at $x=0$ and $x=2$. Now we have our roadmap: from $x=0$ to $x=2$, one curve is the minimum, and from $x=2$ to $x=3$, the other takes over.

By testing a point like $x=1$, we see that $1^2+2=3$ and $2(1)+2=4$. Since $3<4$, the parabola is the lower boundary on the interval $[0,2]$. Conversely, for $x=2.5$, the line is lower.

Our total area $A$ is the sum of two distinct regions:

Let us tackle $A_1$ first. The anti-derivative of $x^2+2$ is $\frac{x^3}{3}+2x$. Evaluating this from $0$ to $2$:

Now for $A_2$, the area under the line from $2$ to $3$. The anti-derivative of $2x+2$ is $x^2+2x$. Evaluating from $2$ to $3$:

Adding these together, we find the total area:

We have arrived at the final step, but do not let your guard down! The question asks for $12A$, not just $A$.

We take our result $A=\frac{41}{3}$ and multiply it by $12$:

And there it is—the elegance of the final answer. The final result is 164. Keep this logic in your toolkit: whenever you see a $\min$ or $\max$ function, find the intersection points first, and the rest will follow.