Answer
=$\frac{7}{2}$
Work Step by Step
$\int^0_1(3x^2+x-5) \ dx$=$-\int^1_0(3x^2+x-5) \ dx$
=$-[3\int^1_0x^2 \ dx+\int^1_0x \ dx-\int^1_05 \ dx$
=$-[3(\frac{1^3}{3}-\frac{0^3}{3})+(\frac{1^2}{2}-\frac{0^2}{2})-5(1-0)]$
=$-(\frac{3}{2}-5)$=$\frac{7}{2}$
