How to convert a Riemann sum to a definite integral

A Riemann sum already contains the pieces of a definite integral: a function value, a rectangle width, and an interval. The job is to identify those pieces inside the sum. Let’s work through an example so you can see where the integrand and both bounds come from.

Match the sum to the integral

For a continuous function on [a, b], equal-width right-endpoint Riemann sums satisfy

\lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}) Δ x = \int_{a}^{b} f (x) d x

Here n is the number of rectangles, and the common width is

Δ x = \frac{b - a}{n}

For right endpoints, the sample point is the left end of the whole interval plus i rectangle widths:

x_{i} = a + i Δ x

The limit is applied to the sum as n approaches infinity. A sum with one fixed, finite n is generally an approximation; its limit gives the integral when the required conditions hold.

Example: Identify the width and sample points

Convert this limit to a definite integral.

\lim_{n \to \infty} \sum_{i = 1}^{n} ((\frac{3 i}{n} - 2)^{2} + 2 (\frac{3 i}{n} - 2) + 7) \frac{3}{n}

The factor outside the polynomial is the rectangle width.

Δ x = \frac{3}{n}

The expression repeated inside the polynomial is the sample point.

x_{i} = - 2 + \frac{3 i}{n}

This matches a + i Δx, with a = −2. The width tells us that the entire interval has length 3.

Find the upper bound

\frac{b - a}{n} = \frac{3}{n}

b - (- 2) = 3

b = 1

So the interval is [−2, 1]. We can also see the last right endpoint by substituting i = n into xᵢ: it gives −2 + 3 = 1.

Find the integrand

Replace each occurrence of the sample point with x. The polynomial in the sum becomes

f (x) = x^{2} + 2 x + 7

We now have the function and the interval. Replace the limit of the sum with integral notation.

\int_{- 2}^{1} (x^{2} + 2 x + 7) d x

That is the requested definite integral. The task here is conversion, so we can stop once the integral is written.

Why the sample-point expression matters

The width gives the length of the interval, while the sample points locate that interval on the x-axis. In this example, $\frac{3}{n}$ gives a length of 3, and the shift by −2 gives the left endpoint. Both pieces are needed to identify the bounds.

Left-endpoint and midpoint sums can also approach the same definite integral for a continuous function. Their sample-point formulas differ, so we match the sampling rule before identifying the interval.

For the notation at the end of the integral, see what dx means in an integral.

Learn mathKrista KingAugust 31, 2016riemann sum, convert, definite integrals