Question

If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to

Hint:

Use integration to find the area of the region bounded by curves.

Answer 1

Correct Answer: 368

1. Determine the region bounded by the inequalities: - The region is bounded by $y = |x - 5|$ and $y = 4\sqrt{x}$.
2. Find the intersection points of the curves: - Solve $y = |x - 5|$ and $y = 4\sqrt{x}$: \[ |x - 5| = 4\sqrt{x} \] - For $x \geq 5$: \[ x - 5 = 4\sqrt{x} \] \[ x - 4\sqrt{x} - 5 = 0 \] - Let $u = \sqrt{x}$, then $u^2 - 4u - 5 = 0$: \[ u = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm 6}{2} \] \[ u = 5 \quad \text{or} \quad u = -1 \quad (\text{not valid}) \] \[ x = 25 \] - For $x<5$: \[ 5 - x = 4\sqrt{x} \] \[ 5 - 4\sqrt{x} - x = 0 \] - Let $u = \sqrt{x}$, then $u^2 + 4u - 5 = 0$: \[ u = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm 6}{2} \] \[ u = 1 \quad \text{or} \quad u = -5 \quad (\text{not valid}) \] \[ x = 1 \]
3. Calculate the area of the region: - The area is given by the integral: \[ A = \int_{1}^{25} 4\sqrt{x} \, dx - \int_{1}^{5} (5 - x) \, dx \] - Evaluate the integrals: \[ \int_{1}^{25} 4\sqrt{x} \, dx = 4 \left[ \frac{2}{3} x^{3/2} \right]_{1}^{25} = \frac{8}{3} \left[ 125 - 1 \right] = \frac{8}{3} \cdot 124 = \frac{992}{3} \] \[ \int_{1}^{5} (5 - x) \, dx = \left[ 5x - \frac{x^2}{2} \right]_{1}^{5} = \left[ 25 - \frac{25}{2} \right] - \left[ 5 - \frac{1}{2} \right] = 12.5 - 4.5 = 8 \] - Total area: \[ A = \frac{992}{3} - 8 = \frac{992 - 24}{3} = \frac{968}{3} = \frac{320}{3} \] - Therefore, $3A = 320$. Therefore, the correct answer is (1) 368.