Set up the equations. We have two curves:
\[ y = 4x - x^2 \]
and
\[ 3y = (x - 4)^2. \]
Rewrite the second equation for \( y \):
\[ y = \frac{(x - 4)^2}{3}. \]
Find the points of intersection. To find the points of intersection, set the two expressions for \( y \) equal to each other:
\[ 4x - x^2 = \frac{(x - 4)^2}{3}. \]
Multiply through by 3 to eliminate the fraction:
\[ 3(4x - x^2) = (x - 4)^2 \] \[ 12x - 3x^2 = x^2 - 8x + 16 \]
Bring all terms to one side of the equation:
\[ -4x^2 + 20x - 16 = 0. \]
Divide by \(-4\):
\[ x^2 - 5x + 4 = 0. \]
Factor the quadratic equation:
\[ (x - 4)(x - 1) = 0. \]
Thus, the points of intersection are \( x = 1 \) and \( x = 4 \).
Set up the integral for the area. The area enclosed by the curves from \( x = 1 \) to \( x = 4 \) is given by the integral of the difference between the upper and lower functions:
\[ \text{Area} = \int_{1}^{4} \left( (4x - x^2) - \frac{(x - 4)^2}{3} \right) dx. \]
Simplify the integrand. Expand \( \frac{(x - 4)^2}{3} \):
\[ \frac{(x - 4)^2}{3} = \frac{x^2 - 8x + 16}{3} = \frac{x^2}{3} - \frac{8x}{3} + \frac{16}{3}. \]
Now rewrite the integrand:
\[ \left( 4x - x^2 - \frac{x^2}{3} + \frac{8x}{3} - \frac{16}{3} \right). \]
Combine like terms:
\[ = \int_{1}^{4} \left( -\frac{4x^2}{3} + \frac{20x}{3} - \frac{16}{3} \right) dx. \]
Integrate term by term. Now integrate each term separately:
\[ \int_{1}^{4} -\frac{4x^2}{3} dx = -\frac{4}{3} \left[ \frac{x^3}{3} \right]_{1}^{4} = -\frac{4}{9} (64 - 1) = -\frac{4 \times 63}{9} = -28. \]
\[ \int_{1}^{4} \frac{20x}{3} dx = \frac{20}{3} \left[ \frac{x^2}{2} \right]_{1}^{4} = \frac{20}{3} \cdot \frac{15}{2} = 50. \]
\[ \int_{1}^{4} -\frac{16}{3} dx = -\frac{16}{3} \cdot (4 - 1) = -\frac{16 \times 3}{3} = -16. \]
Add the results:
\[ \text{Area} = -28 + 50 - 16 = 6. \]
Thus, the answer is:
6
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: