Given: f(x) = casesx, & 0 x

Question

Given: $f(x) = \begin{cases}x, & 0\le x<\dfrac12\ \dfrac12, & x=\dfrac12\ 1-x, & \dfrac12<x\le1\end{cases}$ and $g(x) = \left(x-\dfrac12ight)^2$, $x\in\mathbb R$. Then the area (in sq. units) of the region bounded by the curves $y=f(x)$ and $y=g(x)$ between the lines $2x=1$ and $2x=\sqrt3$, is:

Accepted Answer

On $\left[\dfrac12,\dfrac{\sqrt3}{2}\right]$, $f(x)=1-x$, which lies above $g(x)=\left(x-\dfrac12\right)^2$.

$$\text{Area} = \int_{1/2}^{\sqrt3/2}\left[(1-x)-\left(x-\frac12\right)^2\right]dx$$

$$= \left[x-\frac{x^2}{2}-\frac13\left(x-\frac12\right)^3\right]_{1/2}^{\sqrt3/2} = \frac{\sqrt3}{4}-\frac13$$

Solutions · 1