[tex]\displaystyle\bf\\\boxed{\bf \int\limits^b_a f(x) \, dx=F(b)-F(a)=F(x)~\Bigg |_a^b}~.\\\\\int\limits^2_1 {4x^3+5x^2+6x+3} \, dx = \int\limits^2_1 {4x^3} \, dx + \int\limits^2_1 {5x^2} \, dx + \int\limits^2_1 {6x} \, dx + \int\limits^2_1 {3} \, dx =\\\\x^4~\Bigg |_1^2}+ \frac{5x^3}{3}~\Bigg |_1^2}+3x^2~\Bigg |_1^2}+3x~\Bigg |_1^2}=(16-1)+\bigg(\frac{5\cdot 2^3}{3}-\frac{5\cdot1}{3} \bigg)+\big(3\cdot2^2-3)+(6-3)=\\\\15+\frac{35}{5} +9+3=\boxed{\bf \frac{116}{3}}~.\\\\[/tex]
[tex]\displaystyle\bf\\\int\limits^2_{-1} {-2x^3+4x^2-6x+10} \, dx= -\int\limits^2_{-1} {2x^3} \, dx +\int\limits^2_{-1} {4x^2} \, dx-\int\limits^2_{-1} {6x} \, dx + \int\limits^2_{-1} {10} \, dx=\\\\-\frac{x^4}{2} \Bigg |_{-1}^2 + \frac{4x^3}{3}\Bigg |_{-1}^2 - 3x^2\Bigg |_{-1}^2+10x\Bigg |_{-1}^2=\\\\\bigg (-\frac{16}{2}+\frac{1}{2} \bigg )+\bigg ( \frac{32}{3} - \frac{-4}{3} \bigg ) -\bigg(12-3\bigg)+(20+10)=~\boxed{\bf \frac{51}{2}}~.[/tex]
