JEE PYQ: Area Under Curves Question 14
Question 14 - 2020 (05 Sep Shift 2)
The area (in sq. units) of the region $A = {(x, y) : (x-1)[x] \leq y \leq 2\sqrt{x},\ 0 \leq x \leq 2}$, where $[t]$ denotes the greatest integer function, is:
(1) $\frac{8}{3}\sqrt{2} - \frac{1}{2}$
(2) $\frac{4}{3}\sqrt{2} + 1$
(3) $\frac{8}{3}\sqrt{2} - 1$
(4) $\frac{4}{3}\sqrt{2} - \frac{1}{2}$
Show Answer
Answer: (1) $\frac{8}{3}\sqrt{2} - \frac{1}{2}$
Solution
$[x] = 0$ for $x \in [0,1)$, $[x] = 1$ for $x \in [1,2)$. $A = \int_0^2 2\sqrt{x},dx - \frac{1}{2}(1)(1) = \frac{8\sqrt{2}}{3} - \frac{1}{2}$.
BETA
