Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region, given by the set $\{(x,y)\in \mathbb{R}\times \mathbb{R}\mid x\ge 0,2x^2\le y\le 4-2x\}$ is :

Imagine you are standing on a coordinate plane, looking at a landscape defined by two distinct curves. We have a 'floor' and a 'ceiling'.

The floor is the parabola $y=2x^2$, a steep, upward-opening bowl that starts at the origin. The ceiling is the line $y=4-2x$, a downward-sloping boundary that cuts across the plane.

But there is a gatekeeper here: the constraint $x\ge 0$. This means we are not looking at the entire infinite plane; we are restricted to the right side of the $y$-axis. This is our arena.

Finding the Intersection

To calculate the area, we must know where the floor meets the ceiling. We set the two equations equal:

This is the logical bridge that connects our two curves. Rearranging this gives us the quadratic equation:

Dividing by $2$, we get $x^2+x-2=0$. Factoring this, we find $(x+2)(x-1)=0$.

This gives us two potential intersection points: $x=-2$ and $x=1$. But remember our gatekeeper! The constraint $x\ge 0$ forces us to reject $x=-2$. Thus, the curves meet perfectly at $x=1$.

Summing the Slices

Now, we are ready for the main event. We want to find the area of the region trapped between $x=0$ and $x=1$.

We use the method of vertical strips. Imagine an infinite number of tiny, vertical rectangles of width $dx$ and height equal to the difference between the ceiling and the floor. The height of each strip is $(4-2x)-2x^2$.

To find the total area, we sum these strips using the definite integral:

Integrating term by term, the integral of $4$ is $4x$, the integral of $-2x$ is $-x^2$, and the integral of $-2x^2$ is:

Evaluating this from $0$ to $1$, we get:

This simplifies to $4-1-\frac{2}{3}$, which is $3-\frac{2}{3}$. Finding a common denominator, we get:

We have successfully navigated the geometry, the algebra, and the calculus to arrive at the final answer of $\frac{7}{3}$ square units.