Question Number 159768 by cortano last updated on 21/Nov/21
Commented by Tony6400 last updated on 21/Nov/21
![∫_(≾3) ^4 [6+(x/2)≾(x^2 /2)]dx =[6x+(x^2 /4)≾(x^3 /6)]_(≾3) ^4 =[6(4)+(4^2 /4)≾(4^3 /6)]≾[6(≾3)+(((≾3)^2 )/4)≾(((≾3)^3 )/6)] =[24+4≾((32)/3)]≾[≾18+(9/4)+((27)/6)] =(28≾((32)/3))≾(≾((45)/4)) =((52)/3)+((45)/4)=((343)/(12))sq units](https://www.tinkutara.com/question/Q159776.png)
$$\int_{\precsim\mathrm{3}} ^{\mathrm{4}} \left[\mathrm{6}+\frac{{x}}{\mathrm{2}}\precsim\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]{dx} \\ $$$$=\left[\mathrm{6}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\precsim\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\right]_{\precsim\mathrm{3}} ^{\mathrm{4}} =\left[\mathrm{6}\left(\mathrm{4}\right)+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{4}}\precsim\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{6}}\right]\precsim\left[\mathrm{6}\left(\precsim\mathrm{3}\right)+\frac{\left(\precsim\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{4}}\precsim\frac{\left(\precsim\mathrm{3}\right)^{\mathrm{3}} }{\mathrm{6}}\right] \\ $$$$=\left[\mathrm{24}+\mathrm{4}\precsim\frac{\mathrm{32}}{\mathrm{3}}\right]\precsim\left[\precsim\mathrm{18}+\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{27}}{\mathrm{6}}\right] \\ $$$$=\left(\mathrm{28}\precsim\frac{\mathrm{32}}{\mathrm{3}}\right)\precsim\left(\precsim\frac{\mathrm{45}}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{52}}{\mathrm{3}}+\frac{\mathrm{45}}{\mathrm{4}}=\frac{\mathrm{343}}{\mathrm{12}}{sq}\:{units} \\ $$
Answered by cortano last updated on 21/Nov/21
