Question

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

Answer 1

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

Let,

$F\left(x\right)=\int (4x^3-5x^2+6x+9).dx$

$=4\left(\frac{x^4}{4}\right)-5\left(\frac{x^3}{3}\right)+6\left(\frac{x^2}{2}\right)+9x$

$\Rightarrow F\left(x\right)=x^4-\frac{5x^3}{3}+3x^2+9x$

${\int }_1^{2}94x^2-9x^2+6x+9).dx$

$=F\left(2\right)-F\left(1\right)$

$=(2{)}^4-\frac{5}{3}(2{)}^3+3(2{)}^2+9(2)-(164-\frac{5}{3}(1)63+3(1{)}^2+9)$

$=16-\frac{40}{3}+12+18-[1-\frac{5}{3}+3+9]$

$=\frac{8}{3}+30-[1+\frac{4}{3}+9]$

$=\frac{98}{3}-\left[\frac{34}{3}\right]$

$=\frac{64}{3}$