Question

Integrate:

${\int }_1^2(4x^3-5x^2+6x+9)dx$

Answer 1

Consider the given integral.

$I={\int }_1^2(4x^3-5x^2+6x+9)dx$

$I={[\frac{4x^4}{4}-\frac{5x^3}{3}+\frac{6x^2}{2}+9x]}_1^2$

$I={[x^4-\frac{5x^3}{3}+3x^2+9x]}_1^2$

$I=(2^4-\frac{5\times 2^3}{3}+3\times 2^2+9\times 2)-(1^4-\frac{5\times 1^3}{3}+3\times 1^2+9\times 1)$

$I=(16-\frac{40}{3}+12+18)-(1-\frac{5}{3}+3+9)$

$I=(46-\frac{40}{3})-(13-\frac{5}{3})$

$I=46-\frac{40}{3}-13+\frac{5}{3}$

$I=33-\frac{35}{3}$

$I=\frac{64}{3}$

Hence, this is the answer.