Animated Solution for Mathematics - Definite Integration: Let $\alpha$ be the area of the larger region bounded by the curve $y^2=8x$ and the lines $y=x$ and $x=2$, which lies in the first quadrant. Then the value of $3\alpha$ is equal to ______.

Imagine you are standing on a coordinate plane, looking at the elegant dance between a parabola and a straight line. We are tasked with finding the area of a specific region bounded by the parabola $y^2=8x$ and the line $y=x$, with a vertical divider at $x=2$.

First, we must see the landscape. We have a parabola $y^2=8x$ that opens wide to the right, and a straight line $y=x$ cutting through the origin at a 45-degree angle.

Since we are restricted to the first quadrant, we can express the parabola as $y=\sqrt{8x}=2\sqrt{2}{x}^{1/2}$. This form is much friendlier for the integration that lies ahead.

Before we can measure the area, we must know where the region begins and ends. We equate the two curves: $y^2=8x$ and $y=x$.

Substituting $y=x$ into the parabola equation, we get $x^2=8x$, which simplifies to $x(x-8)=0$. This reveals our intersection points: the origin $(0,0)$ and the point $(8,8)$. These points are the anchors of our region.

To find the total area enclosed by these curves, we use the power of definite integration. We integrate the upper curve minus the lower curve from $x=0$ to $x=8$.

Our integral is:

Performing the integration, we get:

Evaluating this at the limits, we find $A=\frac{128}{3}-32=\frac{32}{3}$. This is the total area of the region trapped between the parabola and the line.

Now, the problem introduces a vertical line $x=2$. Think of this line as a knife slicing through our region, dividing it into two parts: $A_1$ (from $x=0$ to $x=2$) and $A_2$ (from $x=2$ to $x=8$).

We need to find the larger of these two. Let us calculate $A_1$ first:

Evaluating this, we get:

We have the total area $A=\frac{32}{3}$ and the first part $A_1=\frac{10}{3}$. The second part $A_2$ is simply:

Comparing $A_1=\frac{10}{3}$ and $A_2=\frac{22}{3}$, it is clear that $A_2$ is the larger region. Thus, $\alpha =\frac{22}{3}$.

The question asks for $3\alpha$. Multiplying our result by $3$, we get $3\times \frac{22}{3}=22$. The elegance of this final cancellation is the reward for our careful work. You have successfully navigated the geometry and the calculus to arrive at the solution: 22.