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Question Number 171301 by mpakhrur last updated on 12/Jun/22

$${\int }_1^{2}6x^2-2x+3$$

Commented by Kalebwizeman last updated on 12/Jun/22

$$\frac{6x^3}{3}-\frac{2x^2}{2}+3x]\underset{1}{\stackrel{2}{}}$$$$2x^3-x^2+3x]\underset{1}{\stackrel{2}{}}$$$$=[2{\left(2\right)}^3-{\left(2\right)}^2+3\left(2\right)]-[2{\left(1\right)}^3-{\left(1\right)}^2+3\left(1\right)]$$$$=(16-4+6)-(2-1+3)$$$$=18-4$$$$=14$$

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