Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region bounded by the curve $x^2=4y$ and the straight line $x=4y-2$ is :

Welcome, future engineer. Today, we are not just solving a calculus problem; we are exploring the elegant dance between a parabola and a straight line.

When you look at the equations $x^2=4y$ and $x=4y-2$, do not just see symbols. See a physical reality. The first equation, $x^2=4y$, describes a classic, symmetric bowl—an upward-facing parabola with its vertex at the origin.

The second, $x=4y-2$, is a straight line cutting through the Cartesian plane. Our goal is to find the area of the region they enclose together. This is the 'trapped space'—a finite, beautiful slice of the plane.

Before we can calculate the area, we must define the territory. Where do these two curves meet? This is the most critical step in any area-between-curves problem.

We need the intersection points. We have two equations: 1. $4y=x^2$ 2. $4y=x+2$

Since both expressions are equal to $4y$, we can equate them: $x^2=x+2$. This brings us to a simple quadratic equation: $x^2-x-2=0$.

Factoring this is a breeze: $(x-2)(x+1)=0$. This tells us that our curves intersect at $x=2$ and $x=-1$. These are our limits of integration. They are the 'gates' of our region.

Now, we must set up the integral. The area $A$ is defined by the vertical distance between the curves, integrated over the interval $[-1,2]$.

The formula is $A={\int }_{-1}^2(y_{upper}-y_{lower})dx$. As we determined earlier, the line $y=\frac{x+2}{4}$ sits above the parabola $y=\frac{x^2}{4}$ in this interval.

So, our integral becomes:

To make our lives easier, let's pull the constant $\frac{1}{4}$ outside the integral sign. This leaves us with a simple polynomial:

Now, we perform the integration term by term. The integral of $x$ is $\frac{x^2}{2}$, the integral of $2$ is $2x$, and the integral of $x^2$ is $\frac{x^3}{3}$.

Our antiderivative is:

Evaluating at the upper limit ($x=2$):

Evaluating at the lower limit ($x=-1$):

Finally, we subtract the lower limit value from the upper limit value:

The double negative becomes a positive: $\frac{10}{3}+\frac{7}{6}=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$. Multiplying by our constant $\frac{1}{4}$, we get:

There it is: $\frac{9}{8}$ square units. It is a clean, precise result.

Remember, the beauty of JEE problems lies not in the complexity of the arithmetic, but in the clarity of the setup. Always visualize, always find your limits, and always respect the 'upper minus lower' rule. You have mastered this region; now go forth and master the next.