Animated Solution for Mathematics - Definite Integration: Let the area of the bounded region $\{(x,y):0\le 9x\le y^2,y\ge 3x-6\}$ be $A$. Then $6A$ is equal to ________

The problem defines a region bounded by the inequalities $0\le 9x\le y^2$ and $y\ge 3x-6$. These constraints define a specific area on the Cartesian plane.

The condition $9x\le y^2$ implies that we are considering the region outside the parabola $y^2=9x$. The condition $y\ge 3x-6$ places us above the line $y=3x-6$.

To determine the limits of integration, we identify the intersection points of the curves $y^2=9x$ and $y=3x-6$. Substituting the line equation into the parabola equation yields:

Expanding the left side, we obtain:

Rearranging the terms into a standard quadratic form results in:

Dividing the entire equation by $9$, we simplify the expression to:

Factoring the quadratic, we find $(x-1)(x-4)=0$, which gives us the intersection points at $x=1$ and $x=4$.

We focus on the region between $x=0$ and $x=1$. In this interval, the upper boundary is the lower branch of the parabola, $y=-3\sqrt{x}$, and the lower boundary is the line $y=3x-6$.

The area $A$ is calculated using the definite integral:

Substituting the functions into the integral, we get:

This simplifies to:

We perform the integration term by term:

Evaluating at the boundaries $x=1$ and $x=0$:

Thus, the area is $A=\frac{5}{2}$. The problem asks for the value of $6A$:

The final result is 15.