Question

Obtain ${\int }_2^3(3x+8)dx$ as limit of sum.

Answer 1

${\int }_2^3(3x+8)dx=\frac{3x^2}{2}+8x{|}_2^3=\frac{3{\left(3\right)}^2}{2}+8\left(3\right)-\frac{3{\left(2\right)}^2}{2}-8\left(2\right)=\frac{15}{2}+8=\frac{31}{2}$