Animated Solution for Mathematics - Definite Integration: If the area of the region $\{(x,y):1-2x\le y\le 4-x^2,x\ge 0,y\ge 0\}$ is $\frac{\alpha }{\beta }$, $\alpha ,\beta \in \mathbb{N}$, $\text{gcd}(\alpha ,\beta )=1$, then the value of $(\alpha +\beta )$ is :

Welcome, fellow explorer! Today, we are diving into a beautiful problem of area under curves. Imagine standing on the Cartesian plane, looking at the first quadrant.

We have two primary actors: a downward-opening parabola $y=4-x^2$ and a straight line $y=1-2x$. Our goal is to find the area trapped between them, constrained by the axes.

First, let's sketch our curves. The parabola $y=4-x^2$ starts at $(0,4)$ on the $y$-axis and descends to $(2,0)$ on the $x$-axis. It is a smooth, elegant curve.

Now, consider the line $y=1-2x$. It starts at $(0,1)$ on the $y$-axis and hits the $x$-axis at $(0.5,0)$. This line cuts through the first quadrant, creating a small triangular region near the origin.

The region defined by $1-2x\le y\le 4-x^2$ is the area under the parabola but above the line. Since the line is below the parabola in the first quadrant, the total area we seek is the area under the parabola from $x=0$ to $x=2$, minus the area of the small triangle formed by the line and the axes.

This is a classic subtraction trick in calculus.

Let $A_1$ be the area under the parabola. We calculate this using the definite integral:

Integrating $4$ gives $4x$, and integrating $x^2$ gives $\frac{x^3}{3}$. Evaluating from $0$ to $2$, we get:

Now, for the triangle $A_2$, the base is $0.5$ and the height is $1$. So, the area is:

The required area is $A=A_1-A_2=\frac{16}{3}-\frac{1}{4}$. Finding a common denominator of $12$, we get:

We are told this area is $\frac{\alpha }{\beta }$, where $\text{gcd}(\alpha ,\beta )=1$. Thus, $\alpha =61$ and $\beta =12$.

The final sum is $\alpha +\beta =61+12=73$. You have successfully navigated the geometry and the calculus!