Riemann Sums and the Accumulation of Change

Adding up a changing rate

Water flows into a tank at $r(t)$ liters per minute. If the rate were a steady $5$ liters per minute for $4$ minutes, the tank would gain $5 \times 4 = 20$ liters — the area of a rectangle $4$ wide and $5$ tall under the rate’s graph.

When the rate changes, cut the time into short pieces. On each piece the rate is nearly constant, so the water gained is about rate × time. Add up the pieces. The accumulated change is the area under the rate graph, and the sum of thin rectangles approximates it.

Riemann sums

Divide $[a, b]$ into $n$ subintervals of width $\Delta x = \tfrac{b - a}{n}$ with endpoints $x_0 = a, x_1, \dots, x_n = b$. A Riemann sum is

$$\sum_{i=1}^{n} f(x_i^*)\,\Delta x$$

where each $x_i^*$ is a point of the $i$-th subinterval. The common choices name the sum:

Sum	Height of the $i$-th rectangle
left	$f(x_{i-1})$, the left end
right	$f(x_i)$, the right end
midpoint	$f\!\left(\tfrac{x_{i-1} + x_i}{2}\right)$, the middle

The trapezoidal sum replaces each rectangle by a trapezoid; it equals the average of the left and right sums.

• y = x²

Why left and right sums trap the area

Take an increasing function. On each subinterval its smallest value is at the left end and its largest at the right end. So every left rectangle fits under the curve, and every right rectangle reaches above it.

Add them up: the left sum is too small and the right sum is too large. For an increasing function, left sum ≤ area ≤ right sum; for a decreasing function, the order flips. As the pieces get thinner, both sums close in on the area from opposite sides — which is how the exact area is defined in the next lesson.

Sigma notation

$\sum$ means “add up”. The right sum above is

$$\sum_{i=1}^{4} f(x_i)\,\Delta x = \sum_{i=1}^{4} i^2 \cdot 1 = 1 + 4 + 9 + 16 = 30$$

In general, with $x_i = a + i\,\Delta x$, the right sum is $\sum_{i=1}^{n} f(a + i\,\Delta x)\,\Delta x$.

Worked examples

Common mistakes

Practice problems