MHB How Do You Calculate Position from Velocity at Time t?

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To calculate the position from velocity at time t, the velocity function is given as v = sin(πt) with an initial position s(0) = 20. The derivative of the position function, s(t), is equal to the velocity function, leading to the integral s(t) = ∫sin(πt) dt. The integration results in s(t) = -1/π cos(πt) + C, where C is determined using the initial condition. Solving for C gives s(t) = -1/π cos(πt) + 20 + 1/π, confirming that s'(t) equals v(t).
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$\tiny{205.o27.12 }$
$\text{Given the velocity $v=\frac{ds}{dt}$}$
$\text{and the initial position of a body moving along a
coordinate line,}$
$\text{find the body's position at time t}$
\begin{align}
\displaystyle
v&=\sin\left({\pi t}\right) &s(0)&=20 \\
v'&=\cos{(\pi t)} \\
s(t)&=
\end{align}
$\text{lost.. } $
 
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I think it's like this:

$$s(t)=\int\sin(\pi t)\,\text{d}t=-\frac1\pi\cos(\pi t)+C$$

$$s(0)=-\frac1\pi+C=20\implies C=20+\frac1\pi$$

$$s(t)=-\frac1\pi\cos(\pi t)+20+\frac1\pi$$
 
so $$s'(t)=v(t)$$

;)
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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