If $f(x) =\left\{\begin{matrix}x,&0<x<\frac{1}{2}\\\frac{1}{2},&x=\frac{1}{2}\\1-x,&\frac{1}{2}<x<1\end{matrix}\right.$ and $g(x)=\left(x-\frac{1}{2}\right)^2$, then the area bounded by $y = f(x)$ and $y = g(x)$ from $x =\frac{1}{2}$ to $x=\frac{\sqrt{3}}{2}$, is
Answer & explanation
Correct answer: option 1
The area bounded by the curves $y = f(x)$ and $y = g(x)$ is shaded in Fig. Let A be the required area. Then,
$A=\int\limits_{1/2}^{\sqrt{3}/2}(y_2-y_1)dx=\int\limits_{1/2}^{\sqrt{3}/2}\left\{(1-x)-(x-\frac{1}{2})^2\right\}dx$
$⇒A=\left[-\frac{(1-x)^2}{2}-\frac{(x-1/2)^3}{3}\right]_{1/2}^{\sqrt{3}/2}=\frac{\sqrt{3}}{4}-\frac{1}{3}$