Question

Evaluate the following definite integral
${\int }_4^1(x^2-x)dx$

Answer 1

$\begin{array}{c}{\int }_4^1\left(x^2-x\right)dx\left[{\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx\right]\\ ={\int }_1^4-x^{2}dx+{\int }_1^{4}xdx\\ =\frac{-1}{3}{\left[x^3\right]}_1^4+\frac{1}{2}{\left[x^2\right]}_1^4\\ =\frac{-1}{2}\times \left[64-1\right]+\frac{1}{2}\left[16-1\right]\\ =\frac{-1}{3}\times 63+\frac{15}{2}\\ =-21+\frac{15}{2}=\frac{-27}{2}\end{array}$