Answer
$$
f^{\prime}(t)= 2t-3\sin t, \quad f(0)=5
$$
We find that
$$
f(t)= t^{2}+3\cos t+2
$$
Work Step by Step
$$
f^{\prime}(t)= 2t-3\sin t, \quad f(0)=5
$$
The general anti-derivative of $
f^{\prime}(t)= 2t-3\sin t
$ is
$$
f(t)= t^{2}+3\cos t+C
$$
To determine C we use the fact that $f(0)=5$:
$$
f(0)= (0)^{2}+3\cos (0)+C=5
$$
$ \Rightarrow $
$$
3+C=5 \quad \Rightarrow \quad C=2,
$$
so
$$
f(t)= t^{2}+3\cos t+2
$$
