Prove the following

Question:

Given $: f(x)=\left\{\begin{array}{cc}x & 0 \leq x<\frac{1}{2} \\ \frac{1}{2} & x=\frac{1}{2} \\ 1-x & , \quad \frac{1}{2}

and $g(x)=\left(x-\frac{1}{2}\right)^{2}, x \in R$. Then the area

(in sq. units) of the region bounded by the curves, $\mathrm{y}=f(\mathrm{x})$ and $\mathrm{y}=\mathrm{g}(\mathrm{x})$ between the lines, $2 \mathrm{x}=1$ and $2 \mathrm{x}=\sqrt{3}$, is :

  1. $\frac{1}{3}+\frac{\sqrt{3}}{4}$

  2. $\frac{\sqrt{3}}{4}-\frac{1}{3}$

  3. $\frac{1}{2}+\frac{\sqrt{3}}{4}$

  4. $\frac{1}{2}-\frac{\sqrt{3}}{4}$


Correct Option: , 2

Solution:

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