Animated Solution for Mathematics - Definite Integration: The area of the region given by $A=\{(x,y):x^2\le y\le \min \{x+2,4-3x\}\}$ is:

A Journey Through Curves

Imagine you are standing on a vast, flat coordinate plane. Before you lies a landscape defined by three distinct paths.

On the floor, we have the gentle, sweeping curve of the parabola $y=x^2$. Above it, acting as a protective roof, are two straight lines: $y=x+2$ and $y=4-3x$.

Your mission is to calculate the exact area of the region trapped between this parabolic floor and the roof formed by the $\min$ of these two lines. This is an exercise in spatial visualization and the elegance of calculus.

Visualizing the Landscape

To solve this, we must first understand the terrain. The parabola $y=x^2$ is our foundation.

The lines $y=x+2$ and $y=4-3x$ intersect to form a peak, like the ridge of a house. The $\min$ function dictates that the roof is always the lower of the two lines at any given point.

As you walk along the $x$-axis, the roof changes its identity. Our first task is to find where these boundaries meet.

The Intersection Hunt

We need to find the critical points where these curves collide. First, we look at the parabola and the first line: $x^2=x+2$.

Rearranging this, we get the quadratic equation $x^2-x-2=0$. Factoring this, we find $(x-2)(x+1)=0$, giving us intersection points at $x=-1$ and $x=2$. For our region, the left boundary is at $x=-1$.

Next, we look at the parabola and the second line: $x^2=4-3x$. This gives us $x^2+3x-4=0$, which factors to $(x+4)(x-1)=0$. The right boundary of our region is at $x=1$.

Finally, we find the ridge of our roof by setting the two lines equal: $x+2=4-3x$. Solving this, we get $4x=2$, or $x=1/2$. This is the crucial point where the roof switches from one line to the other.

The Calculus of the Roof

Now that we have our boundaries, we can set up our integrals. Because the upper boundary changes at $x=1/2$, we must split the area into two parts, $A_1$ and $A_2$.

For the first part, $A_1$, we integrate from $x=-1$ to $x=1/2$. The upper boundary is $y=x+2$ and the lower is $y=x^2$:

Evaluating this, we find the antiderivative: $\frac{x^2}{2}+2x-\frac{x^3}{3}$. Plugging in the limits:

For the second part, $A_2$, we integrate from $x=1/2$ to $x=1$. Here, the upper boundary is $y=4-3x$:

Evaluating this, the antiderivative is $4x-\frac{3x^2}{2}-\frac{x^3}{3}$. Plugging in the limits:

The Final Sum

Finally, we bring it all together. The total area is the sum of these two parts:

By breaking down the complex 'roof' into manageable segments and applying the power of integration, we have tamed the curves. The final result is $5/3$.