Intuition and the fundamental theorem of calculus I think the following example explains the fundamental theorem of calculus quite intuitively. Or more precisely, that's what I thought; now I'm starting to have some doubts.
Suppose $v(t)$ is the velocity of a car driving along the highway. The units for $t$ are in hours and the units for $v(t)$ are in miles per hour. Assume $v(t)$ is continuous and nonnegative. What is the displacement of the car over one hour (ie., $t \in [0,1]$)? 
Well, if we subdivide $[0,1]$ into $n$ subintervals of equal length, in each subinterval $\left[\frac{k}{n}, \frac{k+1}{n}\right]$ the velocity doesn't change too much for large $n$ and hence can be approximated by $v(\frac{k}{n})$. Therefore, the displacement in $\left[\frac{k}{n}, \frac{k+1}{n}\right]$ is equal to $\frac{1}{n} v(\frac{k}{n}) + \epsilon(k, n)$ where $\epsilon(k, n)$ is a small error dependent on $k$ and $n$. 
Hence $$ \text{Displacement} = x(1) - x(0) = \sum_{k=0}^{n-1}  \frac{1}{n} v\left(\frac{k}{n}\right) + \sum_{k=0}^{n-1} \epsilon(k, n) $$
Note that the above equality holds for all $n$, since we have accounted for the error. If we assume that $ \sum_{k=0}^{n-1} \epsilon(k, n) \to 0$ as $n \to \infty$, then it's easy to see that $$x(1) - x(0) = \lim \sum_{k=0}^{n-1}  \frac{1}{n} v\left(\frac{k}{n}\right) = \int_0^1 v(t) \ dt$$
However, it's not obviously clear to me why $\sum_{k=0}^{n-1} \epsilon(k, n) \to 0$ as $n \to \infty$. Why should this hold intuitively? 

The function $v$ is uniformly continuous on $[0,1]$. Given any $\epsilon > 0$, there is an $N$ such that whenever $n \geq N$, we have, for all $k$ and all $x \in [k/n, (k+1)/n]$, the inequality $|v(x) - v(k/n)| \leq \epsilon$. Thus your error $\epsilon(k,n)$ is bounded in absolute value by $\epsilon/n$. Summing over $k$, the total error in the displacement is bounded in absolute value by $\epsilon$, so long as $n \geq N$.
