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Question Number 159768 by cortano last updated on 21/Nov/21

Commented by Tony6400 last updated on 21/Nov/21

$${\int }_{\precsim 3}^4[6+\frac{x}{2}\precsim \frac{x^2}{2}]dx$$$$={[6x+\frac{x^2}{4}\precsim \frac{x^3}{6}]}_{\precsim 3}^4=[6\left(4\right)+\frac{4^2}{4}\precsim \frac{4^3}{6}]\precsim [6(\precsim 3)+\frac{{(\precsim 3)}^2}{4}\precsim \frac{{(\precsim 3)}^3}{6}]$$$$=[24+4\precsim \frac{32}{3}]\precsim [\precsim 18+\frac{9}{4}+\frac{27}{6}]$$$$=(28\precsim \frac{32}{3})\precsim (\precsim \frac{45}{4})$$$$=\frac{52}{3}+\frac{45}{4}=\frac{343}{12}squnits$$

Answered by cortano last updated on 21/Nov/21

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