Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region $A=\{(x,y)\in R\times R|0<x<3,0<y<4,y<x^2+3x\}$ is :

Welcome, fellow traveler of the coordinate plane. Today, we are not just solving an integral; we are mapping a territory.

Imagine you are standing in the first quadrant of the Cartesian plane. You have a rectangular 'universe' defined by the constraints $0<x<3$ and $0<y<4$. This is our playground, a box of width $3$ and height $4$.

Inside this box, a parabola, defined by the function $y=x^2+3x$, is attempting to grow. It is constrained such that it can only exist where $y<x^2+3x$ and $y<4$. Our mission is to find the area of this specific, bounded region.

Finding the Switching Station

As we trace the parabola $y=x^2+3x$ starting from the origin $(0,0)$, it climbs upward. It is destined to hit the ceiling of our box, the line $y=4$.

To understand where the geometry of our region changes, we must find the exact moment this collision occurs. We set the parabola equal to the ceiling:

Rearranging this into a standard quadratic form, we get $x^2+3x-4=0$. Factoring this, we find $(x+4)(x-1)=0$.

This gives us two potential intersection points: $x=-4$ and $x=1$. However, we must be disciplined. Our domain is strictly $x>0$.

Thus, we discard the negative root and focus entirely on $x=1$. This is our 'switching station.' At $x=1$, the parabola reaches the height of $4$.

For any $x>1$, the parabola would be higher than $4$, but our region is capped at $y=4$. Therefore, the upper boundary of our region is the parabola for $x\in (0,1)$ and the horizontal line $y=4$ for $x\in (1,3)$.

Piecewise Integration

Because the upper boundary changes, we must split our total area $A$ into two distinct parts: $A=A_1+A_2$.

For the first part, $A_1$, we integrate the parabola from $x=0$ to $x=1$:

This is the area of the 'curved sliver' near the origin. Applying the power rule for integration, we get:

Now, we move to the second part, $A_2$. Here, the parabola has 'exited' the box, and the region is now bounded by the flat ceiling $y=4$. We integrate from $x=1$ to $x=3$:

As we discussed, this is simply the area of a rectangle with width $2$ and height $4$. Evaluating the integral confirms this:

We have successfully navigated the two phases of our region. Now, we bring them together to find the total area $A$:

To add these, we find a common denominator:

And there you have it. The area of this region is $\frac{59}{6}$ square units. It is a beautiful result, born from the simple act of recognizing when a curve hits a boundary and knowing how to pivot your strategy.