Answer:
[tex]33\frac{1}{3}[/tex]
Step-by-step explanation:
This is a straightforward evaluation of a definite integral.
[tex]\displaystyle\int_1^5{x^2-x+1}\,dx=\left.\dfrac{x^3}{3}-\dfrac{x^2}{2}+x\right|_1^5=\dfrac{5^3-1^3}{3}-\dfrac{5^2-1^2}{2}+(5-1)\\\\=41\frac{1}{3}-12+4=\boxed{33\frac{1}{3}}[/tex]
