Question Number 171301 by mpakhrur last updated on 12/Jun/22
$$\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{6}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3} \\ $$
Commented by Kalebwizeman last updated on 12/Jun/22
![((6x^3 )/3)−((2x^2 )/2)+3x] _1 ^2 2x^(3 ) −x^2 +3x] _1 ^2 =[2(2)^3 −(2)^2 +3(2)]−[2(1)^3 −(1)^(2 ) +3(1)] =(16−4+6)−(2−1+3) =18−4 =14](https://www.tinkutara.com/question/Q171303.png)
$$\left.\frac{\mathrm{6}{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}\right]\underset{\mathrm{1}} {\overset{\mathrm{2}} {\:}}\: \\ $$$$\left.\mathrm{2}{x}^{\mathrm{3}\:} −{x}^{\mathrm{2}} +\mathrm{3}{x}\right]\underset{\mathrm{1}} {\overset{\mathrm{2}} {\:}} \\ $$$$=\left[\mathrm{2}\left(\mathrm{2}\right)^{\mathrm{3}} −\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2}\right)\right]−\left[\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{1}\right)^{\mathrm{2}\:} +\mathrm{3}\left(\mathrm{1}\right)\right] \\ $$$$=\left(\mathrm{16}−\mathrm{4}+\mathrm{6}\right)−\left(\mathrm{2}−\mathrm{1}+\mathrm{3}\right) \\ $$$$=\mathrm{18}−\mathrm{4} \\ $$$$=\mathrm{14} \\ $$