Show that

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

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

By second fundamental theorem of calculus, we obtain

$I=\mathrm{F}(2)-\mathrm{F}(1)$

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

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

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

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

$=\frac{99-35}{3}$

$=\frac{64}{3}$