The area (in sq units) between the curve $y^2=8 x$ and its latus rectum is

The area (in sq units) between the curve $y^2=8 x$ and its latus rectum is
  1. $\frac{32}{3}$
  2. $\frac{64}{3}$
  3. $\frac{16}{3}$
  4. $\frac{8 \sqrt{2}}{3}$

Solution

The required area $=2 \int_0^2 \sqrt{8 x} d x$ $ =\left.4 \sqrt{2} \frac{2 x^{3 / 2}}{3}\right|_0 ^2=\frac{8 \sqrt{2}}{3}(2 \sqrt{2})=\frac{32}{3} $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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