\(\int\left(2 x^{2}-3 x+5\right) d x=\left[\frac{2 x^{3}}{2}-\frac{3 x^{2}}{2}+5 x\right]_{1}^{2}\)
We subtract the lower limit from the upper limit
\begin{array}{l}
{\left[\frac{2(2)^{3}}{3}-\frac{3(2)^{2}}{2}+5(2)\right]-\left[\frac{2(1)^{3}}{3}-\frac{3(1)^{2}}{2}+5(1)\right]} \\
=\left[\frac{16}{3}-\frac{12}{2}+10\right]-\left[\frac{2}{3}-\frac{3}{2}+\frac{5}{1}\right] \\
{\left[\frac{16}{3}+\frac{4}{1}\right]-\left[\frac{4-9 \times 30}{6}\right]=\left[\frac{16+12}{3}\right]-\left[\frac{25}{6}\right]} \\
\frac{28}{3}-\frac{25}{6}=\frac{56-25}{6}=\frac{31}{6}
\end{array}