Question

Solve;
${\int }_1^2(3x-1)dx$

Answer 1

${\int }_1^2(3x-1)dx$

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

$=[\frac{3(2{)}^2}{2}-2]-[\frac{3(1{)}^2}{2}-1]$

$=[3\times 2-2]-[\frac{3}{2}-1]$

$=6-2-\frac{3}{2}+1$

$=4-\frac{3}{2}+1$

$=\frac{8-3+2}{2}$

$=\frac{7}{2}$.