The Riemann Sum in Sigma Notation

One rectangle, one term

Cut the interval from a to b into n strips of equal width $\Delta x$Δx, and number the strips 1, 2, 3, and so on up to n. In strip number i, pick a sample point and call it $x_i$xᵢ. The rectangle on that strip is $f\left(x_i\right)$f(xᵢ) tall and $\Delta x$Δx wide, so its area is $f\left(x_i\right)\Delta x$f(xᵢ) Δx.

Every rectangle has an area of the same form; only the i changes. So the whole sum is a list of terms $f\left(x_1\right)\Delta x$f(x₁) Δx, $f\left(x_2\right)\Delta x$f(x₂) Δx, $f\left(x_3\right)\Delta x$f(x₃) Δx, and so on up to $f\left(x_n\right)\Delta x$f(xₙ) Δx.

Eight strips under $y=\frac{x^2}{4}$y = x²/4 from x = 0 to x = 4, each $\Delta x=0.5$Δx = 0.5 wide, with each height read at the strip’s right edge. The rectangle on strip i has area $f\left(x_i\right)\Delta x$f(xᵢ) Δx, and the eight add to 6.375.

Sigma adds the list

Sigma, $\sum$Σ, is the instruction to add a list of terms that share one form. The sum $f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\cdots +f\left(x_n\right)\Delta x$f(x₁) Δx + f(x₂) Δx + … + f(xₙ) Δx is written $\sum _{i=1}^{n}f\left(x_i\right)\Delta x$Σ f(xᵢ) Δx from i = 1 to n: put i = 1, then i = 2, and so on up to i = n, into $f\left(x_i\right)\Delta x$f(xᵢ) Δx, and add the results.

The letter i is only a counter. Writing the same sum with r or k in place of i changes nothing.

Where strip i sits

The n strips share the interval equally, so $\Delta x=\left(b-a\right)/n$Δx = (b − a) / n. With a = 1, b = 5 and n = 8, $\Delta x=4/8=0.5$Δx = 4 / 8 = 0.5.

Start at a and step along i whole strips: the point reached is $x_i=a+i\Delta x$xᵢ = a + iΔx. So $x_0=a$x₀ = a, $x_1=a+\Delta x$x₁ = a + Δx, and $x_n=a+n\Delta x=b$xₙ = a + nΔx = b. With a = 1 and $\Delta x=0.5$Δx = 0.5, $x_1=1.5$x₁ = 1.5, $x_2=2$x₂ = 2, and $x_8=5$x₈ = 5.

With i running from 1 to n, the point $a+i\Delta x$a + iΔx is the right edge of strip i, so $\sum _{i=1}^{n}f\left(x_i\right)\Delta x$Σ f(xᵢ) Δx from i = 1 to n is the right sum. The left sum reads the left edges instead, which are $x_0$x₀ up to $x_{n-1}$xₙ₋₁, so it is the same sum with i running from 0 to n − 1. A midpoint sum reads $a+\left(i-\frac{1}{2}\right)\Delta x$a + (i − ½)Δx, for i from 1 to n.

The sum for $y=\frac{x^2}{4}$y = x²/4

On [0, 4] with n strips, $\Delta x=\frac{4}{n}$Δx = 4/n and $x_i=\frac{4i}{n}$xᵢ = 4i/n. The height there is $f\left(x_i\right)={\left(\frac{4i}{n}\right)}^2/4=\frac{4i^2}{n^2}$f(xᵢ) = (4i/n)² / 4 = 4i²/n², so one term is $\frac{4i^2}{n^2}\times \frac{4}{n}=\frac{16i^2}{n^3}$4i²/n² × 4/n = 16i²/n³.

The right sum is therefore $\sum _{i=1}^n\frac{16i^2}{n^3}$Σ 16i²/n³ from i = 1 to n. The $\frac{16}{n^3}$16/n³ is the same in every term, so it comes outside: the right sum is $\frac{16}{n^3}$16/n³ times $\sum _{i=1}^n{i}^2$Σ i² from i = 1 to n. The sum of the first n squares is $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$n(n + 1)(2n + 1)/6, so the right sum is $\frac{16n\left(n+1\right)\left(2n+1\right)}{6n^3}=\frac{8\left(n+1\right)\left(2n+1\right)}{3n^2}$16n(n + 1)(2n + 1)/(6n³) = 8(n + 1)(2n + 1)/(3n²).

Check it at n = 4: $8\times 5\times 9÷\left(3\times 16\right)=\frac{360}{48}=7.5$8 × 5 × 9 ÷ (3 × 16) = 360/48 = 7.5, the right sum found by adding the four rectangles. At n = 8 it gives 8 × 9 × 17 ÷ (3 × 64) = 6.375, and at n = 100 it gives 5.4136.

The left sum runs from i = 0 to n − 1, which adds the squares $0^2$0², $1^2$1², and so on up to ${\left(n-1\right)}^2$(n − 1)². The same formula with n − 1 in place of n gives $\frac{8\left(n-1\right)\left(2n-1\right)}{3n^2}$8(n − 1)(2n − 1)/(3n²): 3.5 at n = 4, and 5.2536 at n = 100.

Let n run to infinity

Write the right sum as $\frac{8}{3}\times \frac{n+1}{n}\times \frac{2n+1}{n}$8/3 × (n + 1)/n × (2n + 1)/n. As n grows without bound, $\frac{n+1}{n}=1+\frac{1}{n}$(n + 1)/n = 1 + 1/n gets as close to 1 as you like, and $\frac{2n+1}{n}=2+\frac{1}{n}$(2n + 1)/n = 2 + 1/n gets as close to 2. So the right sum approaches $\frac{8}{3}\times 1\times 2=\frac{16}{3}$8/3 × 1 × 2 = 16/3. The left sum, $\frac{8}{3}\times \frac{n-1}{n}\times \frac{2n-1}{n}$8/3 × (n − 1)/n × (2n − 1)/n, approaches $\frac{16}{3}$16/3 as well.

That common limit is the area, and it is what the definite integral means: the limit, as n grows without bound, of $\sum f\left(x_i\right)\Delta x$Σ f(xᵢ) Δx is the integral from a to b of f(x) dx. In the limit $\sum$Σ becomes $\int$∫ and $\Delta x$Δx becomes dx, while the height $f\left(x_i\right)$f(xᵢ) becomes f(x). Integrating $\frac{x^2}{4}$x²/4 gives $\frac{x^3}{12}$x³/12, and $\frac{4^3}{12}=\frac{16}{3}$4³/12 = 16/3, the same number.

Sixteen right rectangles under $y=\frac{x^2}{4}$y = x²/4, each 0.25 wide. Their sum is 8 × 17 × 33 ÷ (3 × 256) = 5.84375, closer to $\frac{16}{3}\approx 5.333$16/3 ≈ 5.333 than the 6.375 from eight.

Where the integral sign comes from

The sign $\int$∫ is a stretched S, for sum. The integral from a to b of f(x) dx is read as the sum of f(x) dx, the area of a sliver f(x) tall and dx wide, over every sliver from a to b. Each part of the notation comes from a part of $\sum f\left(x_i\right)\Delta x$Σ f(xᵢ) Δx.

The usual mistakes

Leaving out the width. $\sum f\left(x_i\right)$Σ f(xᵢ) adds heights only. For $y=\frac{x^2}{4}$y = x²/4 in 8 strips the right heights add to 12.75; times the width 0.5, the right sum is 6.375.

Adding the widths alone. $\sum _{i=1}^n\Delta x$Σ Δx from i = 1 to n is $n\times \Delta x=b-a$n × Δx = b − a, the length of the interval, and the curve has dropped out.

Taking $\Delta x$Δx to be b − a, or n. With a = 1, b = 5 and n = 8, $\Delta x$Δx is 4 ÷ 8 = 0.5, not 4 or 8.

Writing $x_i=a+i$xᵢ = a + i. That steps i units along, not i strips; it is right only when $\Delta x=1$Δx = 1.

Reading the limit as a derivative. A derivative is a limit of chord gradients; this is a limit of added-up strips, and it gives the definite integral.

An hourly meter

In the application below, the energy used over a shift is $\sum P\Delta h$Σ P Δh: a load P in kilowatts, read off a table, times a strip $\Delta h=2$Δh = 2 hours wide. The left, right and midpoint sums are the same sum with different sample points.