Question Number 173908 by ali009 last updated on 20/Jul/22
$${find}\:{the}\:{value}\:{of}\:{b}\:{so}\:{that}\:{the}\:{line}\:{y}={b} \\ $$$${divides}\:{the}\:{region}\:{bound}\:{by}\:{the}\:{graphs}\:{of} \\ $$$${the}\:{two}\:{functinos}\:,\:{into}\:{two}\:{regions}\:{of}\:{equal} \\ $$$${area}. \\ $$$${f}\left({x}\right)=\mathrm{9}−{x}^{\mathrm{2}} \:{and}\:{g}\left({x}\right)=\mathrm{0} \\ $$
Commented by mr W last updated on 20/Jul/22
$$\left(\frac{\mathrm{9}−{b}}{\mathrm{9}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\mathrm{9}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}\right)=\frac{\left(\mathrm{2}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)×\mathrm{9}}{\mathrm{2}}\approx\mathrm{3}.\mathrm{33} \\ $$
Commented by mr W last updated on 20/Jul/22
Commented by mr W last updated on 20/Jul/22
Commented by mr W last updated on 21/Jul/22
$$\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }=\left(\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }\right)^{\mathrm{2}} \:\Rightarrow\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }=\left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${A}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}{a}_{\mathrm{1}} {b}_{\mathrm{1}} \\ $$$${A}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}{a}_{\mathrm{2}} {b}_{\mathrm{2}} \\ $$$$\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }=\frac{{a}_{\mathrm{2}} {b}_{\mathrm{2}} }{{a}_{\mathrm{1}} {b}_{\mathrm{1}} }=\left(\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }\right)\left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)=\left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)=\left(\frac{{b}_{\mathrm{2}} }{{b}_{\mathrm{1}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$
Commented by Erikyatusabes last updated on 21/Jul/22
$$ \\ $$$$\left(\frac{\mathrm{9}−{b}}{\mathrm{9}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\mathrm{9}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}\right)=\frac{\left(\mathrm{2}−\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)×\mathrm{9}}{\mathrm{2}}\approx\mathrm{3}.\mathrm{33} \\ $$
Commented by Tawa11 last updated on 21/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$