𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐚𝐫𝐞𝐚 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐢𝐧𝐢𝐭𝐞 𝐫𝐞𝐠𝐢𝐨𝐧 𝐛𝐨𝐮𝐧𝐝𝐞𝐝 𝐛𝐲 𝐭𝐡𝐞 𝐜𝐮𝐫𝐯𝐞 𝐚𝐧𝐝 𝐭𝐡𝐞 𝐗 𝐚𝐱𝐢𝐬 𝐘= 8𝐱-𝐱²-12

Set $Y = 0$:

$$8x - x^2 - 12 = 0$$

Rearrange the terms into standard quadratic form:

$$-x^2 + 8x - 12 = 0$$

Multiply by $-1$ to make the leading coefficient positive:

$$x^2 - 8x + 12 = 0$$

Factor the quadratic equation:

$$(x - 2)(x - 6) = 0$$

This gives us the x-intercepts:

$$x = 2 \quad \text{and} \quad x = 6$$

These are the limits of integration.

The area $A$ is given by the integral of the function from $x=2$ to $x=6$:

$$A = \int_{2}^{6} (8x - x^2 - 12)\,dx$$

First, find the antiderivative of the function:

$$\int (8x - x^2 - 12)\,dx = 8\frac{x^2}{2} - \frac{x^3}{3} - 12x + C$$ $$= 4x^2 - \frac{x^3}{3} - 12x + C$$

Now, evaluate the definite integral using the limits:

$$A = \left[ 4x^2 - \frac{x^3}{3} - 12x \right]_{2}^{6}$$

Substitute the upper limit ($x=6$):

$$F(6) = 4(6)^2 - \frac{(6)^3}{3} - 12(6)$$ $$F(6) = 4(36) - \frac{216}{3} - 72$$ $$F(6) = 144 - 72 - 72$$ $$F(6) = 0$$

Substitute the lower limit ($x=2$):

$$F(2) = 4(2)^2 - \frac{(2)^3}{3} - 12(2)$$ $$F(2) = 4(4) - \frac{8}{3} - 24$$ $$F(2) = 16 - \frac{8}{3} - 24$$ $$F(2) = -8 - \frac{8}{3}$$ $$F(2) = -\frac{24}{3} - \frac{8}{3}$$ $$F(2) = -\frac{32}{3}$$

Finally, subtract $F(2)$ from $F(6)$:

$$A = F(6) - F(2)$$ $$A = 0 - \left(-\frac{32}{3}\right)$$ $$A = \frac{32}{3}$$

The area of the finite region is $\frac{32}{3}$.