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.

Answer 2

We need the area of the region \(R=\{(x,y): |x-5|\le y \le 4\sqrt{x}\}\).

Concept Used:

The region is bounded above by \(y=4\sqrt{x}\) and below by \(y=|x-5|\). The limits of \(x\) are where the curves intersect: \[ |x-5|=4\sqrt{x},\quad x\ge0. \] Set \(x=t^2\) (\(t\ge0\)). Then the two cases give \[ 5-t^2=4t \Rightarrow t=1 \Rightarrow x=1,\qquad t^2-5=4t \Rightarrow t=5 \Rightarrow x=25. \] Hence the region exists for \(x\in[1,25]\), with lower curve \(5-x\) on \([1,5]\) and \(x-5\) on \([5,25]\).

Step-by-Step Solution:

\[ A=\int_{1}^{5}\!\big(4\sqrt{x}-(5-x)\big)\,dx+\int_{5}^{25}\!\big(4\sqrt{x}-(x-5)\big)\,dx. \] Compute: \[ \int 4\sqrt{x}\,dx=\frac{8}{3}x^{3/2},\quad \int (5-x)\,dx=5x-\frac{x^2}{2},\quad \int (x-5)\,dx=\frac{x^2}{2}-5x. \] Thus \[ A=\left[\frac{8}{3}x^{3/2}+\frac{x^2}{2}-5x\right]_{1}^{5}+ \left[\frac{8}{3}x^{3/2}-\frac{x^2}{2}+5x\right]_{5}^{25}. \] This yields \[ A=\frac{8}{3}(5\sqrt5-4)+\frac{40}{3}(10-\sqrt5) = \frac{8}{3}\big(5\sqrt5-4+50-5\sqrt5\big) = \frac{8}{3}\cdot 46=\frac{368}{3}. \]

Final Computation & Result

\[ 3A = \boxed{368}. \]