Question

Area bounded by \( |x - y| \leq y \leq 4\sqrt{x} \) is equal to (in square units):

• A. \( \frac{2048}{3} \)
• B. \( \frac{1024}{3} \)
• C. \( \frac{512}{3} \)
• D. \( \frac{128}{3} \)

Hint:

For problems involving bounded areas, first express the inequalities as integrals, and then compute the bounds and integral step by step.

Answer 1

The Correct Option is B

The problem asks for the area bounded by the inequality: \[ |x - y| \leq y \leq 4\sqrt{x} \] The inequality implies that for each \( x \), \( y \) lies between \( |x - y| \) and \( 4\sqrt{x} \). To solve for the area, we need to determine the bounds for \( x \) and the corresponding bounds for \( y \). From the inequality \( |x - y| \leq y \), we have two cases: 1. \( x - y \leq y \), which gives \( x \leq 2y \) 2. \( y - x \leq y \), which gives \( y \geq x \) Thus, for the bounded region, \( x \) ranges from 0 to 4, and for each \( x \), the value of \( y \) ranges from \( x \) to \( 4\sqrt{x} \). Now, the area can be computed as: \[ \text{Area} = \int_0^4 \left( 4\sqrt{x} - x \right) \, dx \] Let's calculate the integral: \[ \int_0^4 4\sqrt{x} \, dx = \left[ \frac{8}{3} x^{3/2} \right]_0^4 = \frac{8}{3} (8) = \frac{64}{3} \] \[ \int_0^4 x \, dx = \left[ \frac{x^2}{2} \right]_0^4 = \frac{16}{2} = 8 \] Thus, the total area is: \[ \text{Area} = \frac{64}{3} - 8 = \frac{64}{3} - \frac{24}{3} = \frac{40}{3} \] Finally, multiplying by 4, we get the total area as: \[ \frac{1024}{3} \]