The visualization of the proof of this result relies on the interpretation of the $S(t)$ and the Mean Value Theorem for Integrals.
Interpretations:
Area

Accumulation of Change

Suppose $P$ is a continuous function on $[a,b]$. Then the function $S$ defined by $S(t)=\int_a^tP(x)dx$ is interpreted as: the area of the region in the plane enclosed by the segment $[a,t]$ on the Xaxis, the lines $x=a$ and $x=t$, and the graph of $y=P(x)$. The theorem's interpretation of the derivative with graphs says that the slope of the line tangent to the graph of $S$ at $(c,S(c))$ is the "height" $P(c)$ on the graph of $P$.  Suppose $P$ is a continuous function on $[a,b]$. Then the function $S$ defined by $S(t)=\int_a^tP(x)dx$ is interpreted as: the measure of the accumulated change for a quantity $y$ during the interval $[a,t]$ , when $\frac{dy}{dx}=P(x)$ gives the rate of change for $y$ over the interval $[a,b]$. The theorem's interpretation of the derivative with mapping diagrams says that the rate of change of $S$ at $x=c$ is the value of the derivative of $y$, namely, $S'(c)=\frac {dy}{dx}_{x=c}=P(c)$ . 
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Martin Flashman, 9 July 2018, Created with GeoGebra