Question

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

Answer 1

${\int }_1^2(4x^3-5x^2+6x+9)dx$
$=4{\int }_1^2{x}^{3}dx-5{\int }_1^2{x}^{2}dx+6{\int }_1^{2}xdx+9{\int }_1^{2}1dx$
$=4{\left[\frac{x^4}{4}\right]}_1^2-5{\left[\frac{x^3}{3}\right]}_1^2+6{\left[\frac{x^2}{2}\right]}_1^2+9{\left[x\right]}_1^2$
$={\left[x^4\right]}_1^2-5{\left[\frac{x^3}{3}\right]}_1^2+3{\left[x^2\right]}_1^2+9{\left[x\right]}_1^2$
$=(16-1)-\frac{5}{3}(8-1)+3(4-1)+9(2-1)$
$=15-\frac{35}{3}+9+9=33-\frac{35}{3}=\frac{64}{3}$