Algebra Questions

Question Number 146746 by mathdanisur last updated on 15/Jul/21

$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{4}} {\int}}\:\sqrt{\mathrm{16}\:-\:{x}^{\mathrm{2}} }\:{dx}\:=\:? \\$$

Answered by qaz last updated on 15/Jul/21

$$\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{16}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\mathrm{16}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\$$ $$=\mathrm{16}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}\:\theta\mathrm{d}\left(\mathrm{sin}\:\theta\right) \\$$ $$=\mathrm{16}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta\mathrm{d}\theta \\$$ $$=\mathrm{8}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\mathrm{cos}\:\mathrm{2}\theta+\mathrm{1}\right)\mathrm{d}\theta \\$$ $$=\mathrm{8}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta+\theta\right)\mid_{\mathrm{0}} ^{\pi/\mathrm{2}} \\$$ $$=\mathrm{4}\pi \\$$

Commented bymathdanisur last updated on 15/Jul/21

$${thankyou}\:{Ser} \\$$

Answered by liberty last updated on 15/Jul/21

$${let}\:{x}=\mathrm{4sin}\:{t}\:\Rightarrow{dx}=\mathrm{4cos}\:{t}\:{dt} \\$$ $${I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{4cos}\:{t}\:\sqrt{\mathrm{16}−\mathrm{16sin}\:^{\mathrm{2}} {t}}\:{dt} \\$$ $${I}=\mathrm{16}\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{t}\right){dt} \\$$ $${I}=\mathrm{8}\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{t}\right)_{\mathrm{0}} ^{\pi/\mathrm{2}} \\$$ $${I}=\mathrm{8}\left(\frac{\pi}{\mathrm{2}}\right)=\mathrm{4}\pi \\$$

Commented bymathdanisur last updated on 15/Jul/21

$${thankyou}\:{Ser} \\$$

Answered by puissant last updated on 15/Jul/21

$$\mathrm{x}=\mathrm{4cos}\left(\mathrm{u}\right)\:\Rightarrow\:\mathrm{dx}=−\mathrm{4sin}\left(\mathrm{u}\right)\mathrm{du} \\$$ $$\mathrm{K}=\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \sqrt{\mathrm{16}−\mathrm{16cos}^{\mathrm{2}} \left(\mathrm{u}\right)}×\left(−\mathrm{4sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{4sin}\left(\mathrm{u}\right)×\mathrm{4sin}\left(\mathrm{u}\right)\mathrm{du} \\$$ $$=\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}\left(\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2u}\right)}{\mathrm{2}}\right)\mathrm{du} \\$$ $$=\mathrm{8}\left[\mathrm{u}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2u}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{8}\left(\frac{\pi}{\mathrm{2}}\right)=\:\mathrm{4}\pi... \\$$

Commented bymathdanisur last updated on 15/Jul/21

$${thankyou}\:{Ser} \\$$

Answered by alcohol last updated on 15/Jul/21

$${Area}\:{of}\:{a}\:{quarter}\:{circle}\:{of}\:{r}=\mathrm{4} \\$$ $${A}=\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}\pi \\$$

Commented bymathdanisur last updated on 15/Jul/21

$${thanks}\:{Ser} \\$$