Given:

evaluate:

Integral of just the 3 would be $\displaystyle 3x$, and $\displaystyle 6(\frac{1}{3})$ since $\displaystyle x^2dx =$ $\displaystyle \frac{1}{3}$ so

on $\displaystyle [0,1]$ the integral will be $\displaystyle 3(1) + $$\displaystyle 6(\frac{1}{3})$=5? I ommited the subraction of b-a for 3x since f(0)=0.