Let the area enclosed by the lines x+y=2, y=0, x=0 and the curve f(x)= min ²+3/4, 1+[x] where [x] denotes the greatest integer ≤ x, be A. Then the value of 12 ~A is _ _ _ _ _ _ _.

Integral Calculus: JEE Main 2023 Mathematics question

JEE Main 2023 · 8 Apr, Evening shift · Q26

JEE Main 20238 Apr, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

Let the area enclosed by the lines

x+y=2, \mathrm{y}=0, x=0

and the curve

f(x)=\min \left\{x^{2}+\frac{3}{4}, 1+[x]\right\}

where

[x]

denotes the greatest integer

\leq x

, be

\mathrm{A}

. Then the value of

12 \mathrm{~A}

is _____________.

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2023 Integral Calculus Question with Answer (8 Apr Evening shift, Q26) | JEE Arc