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 :

Question

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 :

${1 \over 2} + {{\sqrt 3 } \over 4}$

${1 \over 2} - {{\sqrt 3 } \over 4}$

${1 \over 3} + {{\sqrt 3 } \over 4}$

${{\sqrt 3 } \over 4} - {1 \over 3}$

Solution

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