Riemann Sums and Accumulation

Add many small contributions

Suppose a rate is measured in liters per minute. Over a short interval of width $\Delta t$ , the contribution is approximately rate times duration. Adding these contributions suggests a sum. For a function on $[a,b]$ , a tagged partition gives

\sum_{i=1}^n f(x_i^*)\Delta x_i.

Each sample $x_i^*$ lies in its subinterval. If these sums approach one value as the largest subinterval width tends to zero, independent of the samples, that value is the Riemann integral. Continuous functions on closed bounded intervals satisfy this condition.

Equal-width recipes

With $n$ equal intervals, $\Delta x=(b-a)/n$ and $x_i=a+i\Delta x$ . Left sums use $x_{i-1}$ , right sums use $x_i$ , and midpoint sums use $(x_{i-1}+x_i)/2$ . The sample choice changes a finite approximation, even though all three converge for a continuous function.

For an increasing function, left sums underestimate and right sums overestimate the signed integral. This conclusion comes from interval-wise inequalities, not merely from a picture. Without monotonicity, the same bracketing claim need not hold.

Worked example: four pieces

Approximate $\int_0^1x^2\,dx$ with four intervals. The width is $1/4$ . The right sum is

R_4=\frac14\left[\left(\frac14\right)^2+\left(\frac12\right)^2+\left(\frac34\right)^2+1\right]=\frac{15}{32}.

The left sum is $7/32$ . They bracket the exact value $1/3$ . The midpoint sum is $21/64$ , closer in this example but still not exact.

The sum must include the width. Adding heights alone has the wrong units and grows with the number of samples instead of approaching the accumulated quantity.

Worked example: pass to a limit

For the right sum with $n$ pieces,

R_n=\frac1{n^3}\sum_{i=1}^n i^2=\frac{(n+1)(2n+1)}{6n^2}.

Dividing by $n^2$ and taking the limit gives $1/3$ . The finite error is $1/(2n)+1/(6n^2)$ , which is positive and approaches zero. This calculation explicitly connects the rectangles to the integral rather than using an antiderivative in advance.

Signed contributions

Rectangles below the horizontal axis contribute negative values. The integral is net accumulation; total geometric area uses $|f|$ . If a velocity changes sign, its integral gives displacement, while integrating speed gives distance traveled.

Practice

Find the left sum for $f(x)=x$ on $[0,2]$ with two pieces.
Find the corresponding midpoint sum.
Why must the largest width approach zero for a general partition?

Show worked solutions

Width $1$ , left samples $0,1$ : the sum is $1$ .
Samples $1/2,3/2$ give $2$ , the exact integral of this linear function.
Otherwise a large interval can keep a fixed sampling error even while many tiny intervals are added elsewhere.

Explore

Rectangles and exact area

Try this. Increase the rectangle count and compare left, midpoint and right samples. On this increasing curve, left sums underestimate and right sums overestimate.

Number of rectangles4Sample point

∫₀² x² dx = 8/3 ≈ 2.667. 4 left rectangles: 1.75. Signed error: -0.917. Width Δx = 0.5.

This explorer uses $f(x)=x^2$ on $[0,2]$ , whose exact integral is $8/3$ . Compare its interval and value with the $[0,1]$ worked example above.