Evaluate a definite integral of a polynomial using FTC Part 2

First find an antiderivative of $3x^3 - x + 5$. Using the power rule for antiderivatives, $\int 3x^3\,dx = \frac{3x^4}{4}$, $\int -x\,dx = -\frac{x^2}{2}$, and $\int 5\,dx = 5x$. So $F(x) = \frac{3x^4}{4} - \frac{x^2}{2} + 5x$.

By FTC Part 2, $\int_0^2 (3x^3-x+5)\,dx = F(2) - F(0)$. Compute $F(2) = \frac{3(16)}{4} - \frac{4}{2} + 10 = 12 - 2 + 10 = 20$, and $F(0) = 0$. Therefore the value of the integral is $20 - 0 = 20$.