Question

For what values of $aϵ[0,1]$does the area of the figure bounded by the graph of the function $y=f\left(x\right)$ and the straight lines $x=0,x=1$ & $y=f\left(a\right)$ is at a minimum & for what value it is at a maximum if $f\left(x\right)=\sqrt{1-x^2}$. Find also the maximum & the minimum areas to nearest integer.

Answer 1

Area is minimum when

$a=\frac{0+1}{2}=\frac{1}{2}$

$y\left(a\right)=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{2}}$

$Area={\int }_0^{\frac{1}{2}}\sqrt{1-x^2}dx-{\int }_0^{\frac{1}{2}}\frac{\sqrt{3}}{2}dx+{\int }_{\frac{1}{2}}^1\frac{\sqrt{3}}{2}dx-{\int }_{\frac{1}{2}}^1\sqrt{1-x^2}dx$

$A={[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x]}_0^{\frac{1}{2}}-\frac{\sqrt{3}}{2}{\left[x\right]}_0^{\frac{1}{2}}+\frac{\sqrt{3}}{2}{\left[x\right]}_{\frac{1}{2}}^1-{[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x]}_{\frac{1}{2}}^1$

$=[\frac{1}{4}\ast \frac{\sqrt{3}}{2}+\frac{1}{2}\ast \frac{\pi }{6}]-[\frac{\pi }{4}-[\frac{1}{4}\ast \frac{\sqrt{3}}{2}+\frac{1}{2}\ast \frac{\pi }{6}]]$

$=2[\frac{\sqrt{3}}{8}+\frac{\pi }{12}]-\left[\frac{\pi }{4}\right]=\frac{\sqrt{3}}{4}+\frac{\pi }{6}-\frac{\pi }{4}$

$=\frac{\sqrt{3}}{4}-\frac{\pi }{12}sq.unit$

Area is minimum when a=0, y(0)=1

$Area={\int }_0^{1}1dx-{\int }_0^1\sqrt{1-x^2}dx$

$A=(1{-0)-[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x]}_0^{\frac{1}{2}}$

$=1-\frac{\pi }{4}sq.unit$

and for a=1 y=0, the maximum area value is

$Area={\int }_0^1\sqrt{1-x^2}dx={[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x]}_0^{\frac{1}{2}}=\frac{\pi }{4}$