Integrals and accumulation: adding a continuously varying rate

Partition an interval into small subintervals. In each piece choose a sample point, multiply the function value there by the piece's width, and add the products. For a continuous function on a closed bounded interval, these sums approach a common value as the largest width tends to zero. The sample points need not be left endpoints, although that is a convenient first approximation.

Units clarify the construction. Velocity times a short time interval is a small displacement, so integrating velocity gives displacement units. A density per meter multiplied by a short length gives an amount. The differential symbol records the variable being accumulated over; it should not be ignored as decorative punctuation. Changing from time to position requires transforming the whole integral consistently, a topic developed later.

When a function is positive, its integral contributes positively; negative values contribute negatively. Thus positive and negative portions can cancel. Integrating signed velocity gives net displacement, while integrating speed, the magnitude of velocity, gives total distance traveled. A zero net displacement does not imply no travel. The same distinction separates net electrical flow from total traffic passing in both directions.

A definite integral produces a number for specified bounds, while an indefinite integral describes a family of antiderivatives. Reversing bounds changes the sign of a definite integral. Splitting an interval at an interior point splits its accumulation into two contributions. These properties follow the bookkeeping of oriented accumulation and help catch mistakes without evaluating a complicated formula from scratch.

If a continuous rate is accumulated from a fixed starting point to a moving endpoint, the derivative of that accumulated amount is the current rate. Extending the endpoint a little adds a thin contribution whose average rate approaches the endpoint's value. This is one part of the Fundamental Theorem of Calculus and explains why an area-based construction can be computed using derivatives.

For continuous $f$, find an antiderivative $F$ with $F'=f$ and subtract its endpoint values. Constants disappear in that subtraction. This method is exact under the theorem's assumptions, whereas a finite rectangle sum is generally an approximation. Many useful functions lack a simple symbolic antiderivative, so numerical integration remains important. Existence of an integral and availability of an elementary formula are different questions.