Question

Evaluate ${\int }_1^4\left(x^2-x\right)dx$

Answer 1

${\int }_1^4(x^2-4x)dx$

$={[\frac{x^3}{3}-\frac{x^2}{2}]}_1^4$

$=\frac{1}{3}(64-1)-\frac{1}{2}(16-1)$

$=\frac{63}{3}-\frac{15}{2}$

$=21-\frac{15}{2}$

$=\frac{42-15}{2}=\frac{27}{2}$