Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 134152 by mohammad17 last updated on 28/Feb/21

Commented by mohammad17 last updated on 28/Feb/21

$$helpmesir$$

Answered by bobhans last updated on 28/Feb/21

$$\mathrm{Area}={\int }_1^9(1+\sqrt{\mathrm{x}})-(1+\frac{\mathrm{x}}{3})\mathrm{dx}$$$$={\int }_1^9(\sqrt{\mathrm{x}}-\frac{\mathrm{x}}{3})\mathrm{dx}$$$$={[\frac{2}{3}{\mathrm{x}}^{3/2}-\frac{{\mathrm{x}}^2}{6}]}_1^9$$$$=(\frac{2}{3}.(27-1)-\frac{1}{6}\left(80\right))$$$$=\frac{52}{3}-\frac{40}{3}=4$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com