Integration Questions

Question Number 95653 by rb222 last updated on 26/May/20

$${arc}\:{length}\:\mathrm{3}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{1}\: \\$$$${from}\:{x}=\mathrm{0}\:{and}\:{x}=\mathrm{1}\: \\$$$${help}\:{please}\:{sir} \\$$

Answered by john santu last updated on 26/May/20

$$\mathrm{Arc}\:\mathrm{L}\:=\:\underset{\mathrm{0}\:} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} }\:\mathrm{dx}\: \\$$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{9}}{\mathrm{2}}\sqrt{\mathrm{x}}\: \\$$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\sqrt{\mathrm{1}+\frac{\mathrm{81x}}{\mathrm{4}}}\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{4}+\mathrm{81x}}\:\mathrm{dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\left(\mathrm{4}+\mathrm{81x}\right)^{\mathrm{3}/\mathrm{2}} }{\frac{\mathrm{243}}{\mathrm{2}}}\:\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$=\frac{\mathrm{85}\sqrt{\mathrm{85}\:}\:−\mathrm{8}\:}{\mathrm{243}}\: \\$$

Answered by MAB last updated on 26/May/20

$${dl}=\sqrt{\left({dx}\right)^{\mathrm{2}} +\left({dy}\right)^{\mathrm{2}} } \\$$$${dl}=\sqrt{\mathrm{1}+\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} }{dx} \\$$$${l}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\left({f}\:'\left({x}\right)\right)^{\mathrm{2}} }{dx} \\$$$${l}=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\frac{\mathrm{81}}{\mathrm{4}}{x}}{dx} \\$$$${l}=\left[\frac{\mathrm{8}}{\mathrm{243}}\left(\mathrm{1}+\frac{\mathrm{81}}{\mathrm{4}}{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$${l}=\frac{\mathrm{85}\sqrt{\mathrm{85}}−\mathrm{8}}{\mathrm{243}} \\$$

Commented by rb222 last updated on 27/May/20

$${thanks}\:{sir} \\$$

Commented by MAB last updated on 06/Jun/20

$${you}\:{are}\:{welcome} \\$$