Animated Solution for Mathematics - Definite Integration: Let the area of the region enclosed by the curves $y=3x,2y=27-3x$ and $y=3x-x\sqrt{x}$ be $A$. Then $10A$ is equal to

Before we can calculate the area, we must understand the shape of the region defined by the three curves: $C_1:y=3x$ $C_2:y=\frac{27-3x}{2}$ $C_3:y=3x-x\sqrt{x}$

To find the corners of this region, we calculate the intersection points. First, for $C_1$ and $C_2$: Plugging $x=3$ into $y=3x$, we find the first corner at $(3,9)$.

Next, for $C_1$ and $C_3$: This gives us the second corner at $(0,0)$.

Finally, for $C_2$ and $C_3$: Substituting $t=\sqrt{x}$, we solve $2t^3-9t^2+27=0$. Testing values, we find $t=3$ is a root, so $\sqrt{x}=3$, which means $x=9$. Our final corner is $(9,0)$.

With vertices at $(0,0)$, $(3,9)$, and $(9,0)$, we observe that the upper boundary changes at $x=3$. From $x=0$ to $x=3$, the upper curve is $C_1$, while from $x=3$ to $x=9$, the upper curve is $C_2$. The lower boundary remains $C_3$ throughout.

We must split the total area $A$ into two distinct integrals:

First, we simplify the integrands. For the first interval:

For the second interval:

Now, we perform the integration for the first part:

For the second part, we integrate term by term:

Evaluating the second integral at the boundaries: Upper limit ($x=9$): $\frac{243}{2}-\frac{729}{4}+\frac{486}{5}$ Lower limit ($x=3$): $\frac{81}{2}-\frac{81}{4}+\frac{18\sqrt{3}}{5}$

When we combine the two areas, the $\frac{18\sqrt{3}}{5}$ terms cancel out. We are left with:

The final requested value is $10A$: