Answer:
[tex]\frac{53}{6}[/tex]
Step-by-step explanation:
[tex]\int\limits^4_3 {x^2-x} \, dx[/tex]
= [ [tex]\frac{x^3}{3}[/tex] - [tex]\frac{x^2}{2}[/tex] ] ← evaluate for upper limit - lower limit
= ( [tex]\frac{64}{3}[/tex] - [tex]\frac{16}{2}[/tex] ) - ( [tex]\frac{27}{3}[/tex] - [tex]\frac{9}{2}[/tex] )
= [tex]\frac{64}{3}[/tex] - 8 - 9 + [tex]\frac{9}{2}[/tex]
= [tex]\frac{155}{6}[/tex] - 17
= [tex]\frac{53}{6}[/tex]