Integral is area under the graph Can somebody explain why integral is an area under the graph of function of integral, not the area under the function that will be calculated after the integration of the function?
 A: What about thinking about it in terms of distance $s(t)$ and velocity $v(t)$? Distance $s(t)$ is the integral of velocity $v(t)$.
The easiest case is when velocity is constant, then the distance can be obtained by $v(t)\cdot \delta t$, which is exactly the area under $v(t)$ (the area of the rectangle length times width).
If the velocity is not constant, you assume it is constant in a small time period. You still can obtain the distance by finding the area of the small rectangles, and add them up.
In summary, distance $s(t)$ is the area under the velocity function $v(t)$. 
A: 
The figure is an help to intuitively understand the fundamental theorem of calculus.
The area under the graph, starting from a point (here $x=0$) ,is a function $F(x)$. Passing from $x$ to $x+dx$ the area increments (approximately) by $dF=f(x)dx$ that is the area of the pink rectangle.
If $dx \rightarrow 0$ we have $\lim_{x \rightarrow 0}\dfrac{dF(x)}{dx}=\lim_{x \rightarrow 0}\dfrac{F(x+dx)-F(x)}{dx} =f(x). I.e. $f(x) is the derivative of $F(x)$.
Obviously this is not a rigorous proof, but give the right intuition.
