Answer
$f'(t)=\frac{t^{2}}{2}-\sin(t)+2 $
$f(t)=\frac{t^{3}}{6}+\cos(t)+2t-3 $
Work Step by Step
$$f''(t)=t-\cos(t)$$
Integrate both sides with respect to $t$:
$$\int f''(t)dt=\int (t-\cos(t))dt$$
$$f'(t)=\frac{t^{2}}{2}-\sin(t)+c$$
Find $c$.
$$f'(0)=\frac{0^{2}}{2}-\sin(0)+c$$
$$2=\frac{0^{2}}{2}-\sin(0)+c$$
$$2=c$$
$$f'(t)=\frac{t^{2}}{2}-\sin(t)+c \to f'(t)=\frac{t^{2}}{2}-\sin(t)+2 $$
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$$f'(t)=\frac{t^{2}}{2}-\sin(t)+2 $$
Integrate both sides with respect to $t$:
$$\int f'(t)dt=\int (\frac{t^{2}}{2}-\sin(t)+2)dt $$
$$f(t)=\frac{t^{3}}{6}+\cos(t)+2t+k $$
Find $k$.
$$f(0)=\frac{0^{3}}{6}+\cos(0)+2\cdot 0+k $$
$$-2=\frac{0^{3}}{6}+\cos(0)+2\cdot 0+k $$
$$-2=0+1+0+k $$
$$k=-3 $$
$$f(t)=\frac{t^{3}}{6}+\cos(t)+2t+k \to f(t)=\frac{t^{3}}{6}+\cos(t)+2t-3 $$
