Theorem CIS.FTC.I. Fundamental Theorem of Calculus I. (Derivative Form):
Suppose  $P$ is a continuous function on an interval, $I$, with $a$ any number in that interval. For any number $t\in I$, define the function $S$ by $S(t)=\int_a^tP(x)dx$. Then for any $c\in I$, $S'(c)=P(c)$.

Comment: The Fundamental Theorem of Calculus I in its derivative form and its proof are usually visualized using the area interpretation of the definite integral, connecting the definite integral with measuring "area" of regions in the plane. Using mapping diagrams in the visualization of this result connects the definite integral with the accumulation of change.
Interpretations:
 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 X-axis, 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 change for a quantity $y$ during the interval $[a,t]$ , when  $\frac{dy}{dx}=P(x)$ for the interval $[a,b]$. The theorem's interpretation of the derivative with mapping diagrams says that the rate of change of  $S$  at time $c$ is the value of the derivative of $y$, namely, $\frac {dy}{dx}|_{x=c}=P(c)$ .