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Question: Lk2If f(x) = \(\left\{ \begin{array} { l c } \mathrm { x } ; & 0 \leq \mathrm { x } < 1 / 2 \\ 1 / ...

Lk2If f(x) = {x;0≤x<1/21/2;x=1/21−x;1/2<x≤1\left\{ \begin{array} { l c } \mathrm { x } ; & 0 \leq \mathrm { x } < 1 / 2 \\ 1 / 2 ; & \mathrm { x } = 1 / 2 \\ 1 - \mathrm { x } ; & 1 / 2 < \mathrm { x } \leq 1 \end{array} \right. and g(x) = (x – 1/2)2, x Î R. Then the area of the portion bounded between g(x) and f(x) in the interval [1/2, 3\sqrt { 3 }/2] is

A

3/4\sqrt { 3 } / 4 – 1/3

B

3/4\sqrt { 3 } / 4 + 1/3

C

0

D

3/12\sqrt { 3 } / 12

Answer

3/4\sqrt { 3 } / 4 – 1/3

Explanation

Solution

In the given interval given curves meet at x = 3/2\sqrt { 3 } / 2 So required area

= ∫1/23/2{f(x)−g(x)}dx\int _ { 1 / 2 } ^ { \sqrt { 3 } / 2 } \{ f ( \mathrm { x } ) - \mathrm { g } ( \mathrm { x } ) \} \mathrm { dx }

= [x−x22−(x−1/2)3]1/23/2\left[ x - \frac { x ^ { 2 } } { 2 } - \frac { ( x - 1 / 2 ) } { 3 } \right] _ { 1 / 2 } ^ { \sqrt { 3 } / 2 } = 34−13\frac { \sqrt { 3 } } { 4 } - \frac { 1 } { 3 }