Integral Calculus: JEE Main 2020 Mathematics question

JEE Main 2020 · 9 Jan, Evening shift · Q2827

JEE Main 20209 Jan, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

Given :

f(x) = \left\{ {\matrix{ {x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr {{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr {1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr } } \right.

and

g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R

Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x =

\sqrt 3

, is :

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JEE Main 2020 Integral Calculus Question with Solution (9 Jan Evening shift, Q2827) | JEE Arc

JEE Main 2020 · 9 Jan, Evening shift · Q2827

JEE Main 20209 Jan, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

Given :

f(x) = \left\{ {\matrix{ {x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr {{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr {1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr } } \right.

and

g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R

Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x =

\sqrt 3

, is :