Calculus (3rd Edition)

$$f(t) =\frac{1}{2}t^2- \frac{1}{3}t^3-t+2$$
Given$$f^{\prime \prime}(t)=1-2 t, f(0)=2, \text { and } f^{\prime}(0)=-1$$ Then \begin{align*} f'(t) &= \int (1-2t)dt\\ &= t-t^2+c \end{align*} At $t= 0$, $f'(0)=-1,$ then $c= -1$, hence $$f'(t)= t-t^2-1$$ \begin{align*} f(t) &= \int ( t-t^2-1 )dt\\ &= \frac{1}{2}t^2- \frac{1}{3}t^3-t+c_1 \end{align*} At $t=0$ $f=2$, then $c_1= c_1=2$, hence $$f(t) =\frac{1}{2}t^2- \frac{1}{3}t^3-t+2$$