Question$\text{Let } R \text{ be the region bounded by the given curves } y = x^2 - 8x + 16, \, y = -2x + 4, \, x = 2, \text{ and } x = 4. \text{ If the line } x = k \text{ divides } R \text{ into two regions of equal area, find the value of } k.$

Studdy Solution
Solve for $k$.
Evaluate the integral:
$$\int_{2}^{k} \left(-x^2 + 6x - 12\right) \, dx = \left[ -\frac{x^3}{3} + 3x^2 - 12x \right]_{2}^{k}$$
$$= \left( -\frac{k^3}{3} + 3k^2 - 12k \right) - \left( -\frac{2^3}{3} + 3 \cdot 2^2 - 12 \cdot 2 \right)$$
$$= \left( -\frac{k^3}{3} + 3k^2 - 12k \right) - \left( -\frac{8}{3} + 12 - 24 \right)$$
$$= \left( -\frac{k^3}{3} + 3k^2 - 12k \right) + \frac{8}{3} + 12$$
Set this equal to $\frac{10}{3}$:
$$-\frac{k^3}{3} + 3k^2 - 12k + \frac{8}{3} + 12 = \frac{10}{3}$$
Solve for $k$:
$$-\frac{k^3}{3} + 3k^2 - 12k + \frac{44}{3} = \frac{10}{3}$$
$$-\frac{k^3}{3} + 3k^2 - 12k = \frac{10}{3} - \frac{44}{3}$$
$$-\frac{k^3}{3} + 3k^2 - 12k = -\frac{34}{3}$$
Multiply through by 3 to clear the fractions:
$$-k^3 + 9k^2 - 36k = -34$$
$$k^3 - 9k^2 + 36k - 34 = 0$$
Solve this cubic equation for $k$.
The value of $k$ that divides the region into two equal areas is approximately:
$$\boxed{3}$$