Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 45

$s(t)=-\dfrac{\cos \pi t}{\pi}+\dfrac{1}{\pi}$

Here, $v= \sin \pi t$ and $\dfrac{d}{dt}(-\dfrac{\cos \pi t}{\pi}) =\sin \pi t$ and thus, the general form of the function is: $s(t)=-\dfrac{\cos \pi t}{\pi}+C$ Since we have $s(0)=0$ Therefore, $s(0)=0 \implies 0=-\dfrac{\cos \pi (0)}{\pi}+C$ Thus, $C=\dfrac{1}{\pi}$ Hence, $s(t)=-\dfrac{\cos \pi t}{\pi}+\dfrac{1}{\pi}$