Question

The area of the region bounded by the curve $y=x^2$ and the line $y=16$ is:

• (A) $\frac{32}{3}$ sq.units
• (B) $\frac{256}{3}$ sq.units
• (C) $\frac{64}{3}$ sq.units
• (D) $\frac{128}{3}$ sq.units

Answer 1

The Correct Option is B

Explanation:
Given equation of curves are$y=x^2\cdots (1)$$y=16\cdots (2)$By solving both equation $(1)$ and $(2)$,$x^2=16$$x=4,-4$$\therefore$ Points of intersection are $(4,16)$ and $(-4,16)$.

From the figure we have,Required Area $={\int }_{-4}^4(16-x^2)dx$By using Integral property,$A=2{\int }_0^4(16-x^2)dx$$=2{[16x-\frac{x^3}{3}]}_0^4$$=2{[16x-\frac{x^3}{3}]}_0^4$$=2[64-\frac{64}{3}]$$=2\times 64\times \frac{2}{3}$$A=\frac{256}{3}$ sq.unitsHence, the correct option is (B).