Chapter 4 - Review - Exercises - Page 361: 69

$f(t) = t^2+3~cos~t+2$

$f'(t) = 2t-3~sin~t$ $f(t) = \int (2t-3~sin~t)~dt = t^2+3~cos~t+C$ Note that $f(0) = 5$ We can find the value of $C$: $f(0) = (0)^2+3~cos~(0)+C = 5$ $0+3+C = 5$ $C = 2$ Therefore: $f(t) = t^2+3~cos~t+2$