Riemann Sums, Sigma Notation and the…

Part 1 · Hook

Why this matters

A Riemann sum with four rectangles is a rough guess. With forty it is good; with four thousand, a computer cannot tell it from the true area. Mathematicians asked what happens when the number of rectangles goes to infinity, and the answer has its own name and its own symbol: the definite integral.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the width of each subinterval when $[1, 4]$ is split into 6 equal pieces?

0.5
6
4/6
3

Show the answer

Width $= \frac{4 - 1}{6} = 0.5$ .

Correct: 0.5:
6:
4/6:
3:

2. What is $\lim_{n \to \infty} \frac{3}{n}$ ?

0
3
it does not exist
1

Show the answer

As n grows without bound, 3/n shrinks toward 0.

Correct: 0:
3:
it does not exist:
1:

Part 4 · See it

See it first

Three panels show right Riemann sums for f(x) = 0.25x squared + 1 on [0, 4] with n = 4, 8 and 32 rectangles: 11.5, 10.375 and about 9.586. The rectangles hug the curve more closely each time and the sums approach the definite integral, 28/3, about 9.333. — Right Riemann sums with 4, 8 and 32 rectangles close in on the definite integral. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Use $n$ equal subintervals of $[a, b]$ each is $\Delta x = \frac{b - a}{n}$ wide, with right ends $x_i = a + i\,\Delta x$
Write the Riemann sum in sigma notationit becomes $\sum_{i=1}^{n} f(x_i)\,\Delta x$
Let $n \to \infty$ the sums approach one number, the definite integral $\int_a^b f(x)\,dx$
Read a limit of sumsthe width gives $b - a$ , the points give $a$ , and the height gives $f$

Part 6 · Key ideas

Key ideas

$\int_a^b f(x)\,dx$ $= \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x$ with $\Delta x = \frac{b - a}{n}$ and $x_i = a + i\,\Delta x$ .
In $\int_a^b f(x)\,dx$ , $f(x)$ is the integrand, $a$ and $b$ are the limits of integration, and $dx$ names the variable.
For a positive function, the definite integral is the area under the graph; for a rate, the accumulated change, in the accumulated units.
A calculator finds definite integrals numerically: write the integral as your setup, then the value to three decimal places.

Part 7 · Misconception

A common mistake

The wrong idea: In $\lim_{n \to \infty} \sum \frac{3}{n}\left(2 + \frac{3i}{n}\right)$ , the function is $f(x) = 2 + x$ .

What actually happens: The points $x_i = 2 + \frac{3i}{n}$ already include the 2: they run over $[2, 5]$ , and the height at $x_i$ is $x_i$ . So the limit is $\int_2^5 x\,dx$ . (It also equals $\int_0^3 (2 + x)\,dx$ , where $x$ runs over $[0, 3]$ : the interval and the function must be read together.)

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A limit of Riemann sums

Let

S_n

= \sum_{i=1}^{n} \frac{3}{n}\left(2 + \frac{3i}{n}\right)

and

L = \lim_{n \to \infty} S_n

. The table shows

S_n

for a few values of

n

.

The sum S_n for growing n
$n$	1	10	100	1000
$S_n$	15	10.95	10.545	10.5045

1. In the sum, what interval is being split up, and into how many subintervals?

$[2, 5]$ , into $n$ subintervals of width $\frac{3}{n}$
$[0, 3]$ , into $n$ subintervals of width $\frac{3}{n}$
$[2, 5]$ , into 3 subintervals of width $\frac{1}{n}$
$[0, 1]$ , into $n$ subintervals of width $\frac{1}{n}$

Show the answer

$\Delta x = \frac{3}{n}$ means the interval has length 3. The points are $x_i = 2 + i\Delta x$ , so the interval starts at 2 and ends at $2 + 3 = 5$ .

Correct: $[2, 5]$ , into $n$ subintervals of width $\frac{3}{n}$ : Correct: the points $2 + \frac{3i}{n}$ run from just above 2 (at $i = 1$ ) up to 5 (at $i = n$ ), and each step is $\frac{3}{n}$ .
$[0, 3]$ , into $n$ subintervals of width $\frac{3}{n}$ : The points start at 2, not 0: $x_i = 2 + \frac{3i}{n}$ . The width 3 is right, but the interval is shifted.
$[2, 5]$ , into 3 subintervals of width $\frac{1}{n}$ : The number of terms is $n$ (the sum runs $i = 1$ to $n$ ), and the factor $\frac{3}{n}$ is the width.
$[0, 1]$ , into $n$ subintervals of width $\frac{1}{n}$ : The width is $\frac{3}{n}$ , so the interval is 3 long, and the points start at 2.

2. Which definite integral equals $L$ ?

$\int_2^5 (2 + x)\,dx$
$\int_0^1 (2 + 3x)\,dx$
$\int_0^3 x\,dx$
$\int_2^5 x\,dx$

Show the answer

Match the pattern $\sum f(x_i)\,\Delta x$ : $\Delta x = \frac{3}{n}$ and $x_i = 2 + \frac{3i}{n}$ run over $[2, 5]$ , and $f(x_i) = x_i$ . So $L$ $= \int_2^5 x\,dx$ .

$\int_2^5 (2 + x)\,dx$ : The 2 is already inside $x_i = 2 + \frac{3i}{n}$ . With $x$ running from 2 to 5, the height is just $x$ ; adding 2 counts it twice.
$\int_0^1 (2 + 3x)\,dx$ : This matches the heights with $x = \frac{i}{n}$ , but then the width would be $\frac{1}{n}$ . The factor $\frac{3}{n}$ makes this integral 3 times too small.
$\int_0^3 x\,dx$ : If $x$ ran from 0 to 3, the heights would be $\frac{3i}{n}$ . Here they are $2 + \frac{3i}{n}$ .
Correct: $\int_2^5 x\,dx$ : Correct: $\Delta x = \frac{3}{n}$ , $x_i = 2 + \frac{3i}{n}$ , and the height $f(x_i)$ is $x_i$ itself, so $f(x) = x$ on $[2, 5]$ .

3. Which kind of Riemann sum is inside the limit?

a midpoint Riemann sum
a trapezoidal sum
a right Riemann sum
a left Riemann sum

Show the answer

$x_i = 2 + i\cdot\frac{3}{n}$ for $i = 1, \ldots, n$ are the right ends of the subintervals $[x_{i-1}, x_i]$ . In the limit, left and right sums give the same value.

a midpoint Riemann sum: Midpoints would be $2 + \frac{3(i - 1/2)}{n}$ . These points are whole steps from 2.
a trapezoidal sum: A trapezoidal sum averages two heights on each subinterval; each term here has a single height.
Correct: a right Riemann sum: Correct: at $i = n$ the point is $2 + 3 = 5$ , the right end; at $i = 1$ it is the right end of the first subinterval.
a left Riemann sum: A left sum would start at $x_0 = 2$ (from $i = 0$ ) and stop before 5. This sum reaches 5 at $i = n$ .

4. The table suggests $L = 10.5$ . Which region has area $L$ ?

the rectangle 3 wide and 5 tall over $[2, 5]$
the trapezoid under $y = x$ from $x = 2$ to $x = 5$
the triangle under $y = x$ from $x = 0$ to $x = 3$
the region under $y = 2 + x$ from $x = 2$ to $x = 5$

Show the answer

$L$ $= \int_2^5 x\,dx$ , the area under $y = x$ between 2 and 5: a trapezoid with parallel sides 2 and 5 and width 3, area $\frac{21}{2} = 10.5$ , matching the table.

the rectangle 3 wide and 5 tall over $[2, 5]$ : That rectangle (area 15) is the single right rectangle $S_1$ . The limit fills only the region under the line.
Correct: the trapezoid under $y = x$ from $x = 2$ to $x = 5$ : Correct: $L$ $= \int_2^5 x\,dx$ , and that trapezoid has area $\frac{2 + 5}{2} \cdot 3 = 10.5$ .
the triangle under $y = x$ from $x = 0$ to $x = 3$ : That region has area 4.5. The points $x_i$ run over $[2, 5]$ , not $[0, 3]$ .
the region under $y = 2 + x$ from $x = 2$ to $x = 5$ : That counts the 2 twice: its area is $\frac{4 + 7}{2} \cdot 3 = 16.5$ .

5. Which limit equals $\int_1^3 x^3\,dx$ ?

$\lim_{n \to \infty} \sum_{i=1}^{n} \left(1 + \frac{2i}{n}\right)^3 \frac{2}{n}$
$\lim_{n \to \infty} \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^3 \frac{2}{n}$
$\lim_{n \to \infty} \sum_{i=1}^{n} \left(1 + \frac{3i}{n}\right)^3 \frac{3}{n}$
$\lim_{n \to \infty} \sum_{i=1}^{n} \left(1 + \frac{2i}{n}\right)^3$

Show the answer

For $\int_a^b f(x)\,dx$ : $\Delta x = \frac{b - a}{n} = \frac{2}{n}$ and $x_i = a + i\Delta x = 1 + \frac{2i}{n}$ , so the sum is $\sum f(x_i)\Delta x$ with $f(x) = x^3$ .

Correct: $\lim_{n \to \infty} \sum_{i=1}^{n} \left(1 + \frac{2i}{n}\right)^3 \frac{2}{n}$ : Correct: $\Delta x = \frac{3 - 1}{n} = \frac{2}{n}$ and $x_i = 1 + \frac{2i}{n}$ .
$\lim_{n \to \infty} \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^3 \frac{2}{n}$ : These points run over $[0, 2]$ . The interval starts at 1, so $x_i = 1 + \frac{2i}{n}$ .
$\lim_{n \to \infty} \sum_{i=1}^{n} \left(1 + \frac{3i}{n}\right)^3 \frac{3}{n}$ : The width uses the upper limit alone. $\Delta x = \frac{b - a}{n} = \frac{2}{n}$ .
$\lim_{n \to \infty} \sum_{i=1}^{n} \left(1 + \frac{2i}{n}\right)^3$ : Each term needs its width $\Delta x = \frac{2}{n}$ ; without it the sum grows without bound.

6. Let $h(x) = -2$ for all $x$ . What is $\int_1^4 h(x)\,dx$ ?

$-6$
$6$
$-8$
$-2$

Show the answer

Any Riemann sum is $\sum (-2)\Delta x_i$ $= -2 \cdot (4 - 1)$ $= -6$ , so the limit is $-6$ . A definite integral of a negative function is negative.

Correct: $-6$ : Correct: every Riemann sum on [1, 4] is the sum of (negative 2) times widths that add to 3.
$6$ : The size of the rectangle is 6, but each product in a Riemann sum is a width times a height of negative 2: each one is negative.
$-8$ : The width of the interval is 4 - 1 = 3, not 4.
$-2$ : That is the height. The integral is height times width, (negative 2) times 3.

Part 9 · Summary

Summary

The definite integral

\int_a^b f(x)\,dx

is the limit of Riemann sums

\sum f(x_i)\,\Delta x

as

n \to \infty

, with

\Delta x = \frac{b - a}{n}

and

x_i = a + i\,\Delta x

. Translate both ways by matching the width, the points and the height. For a rate, the integral is the accumulated change.

Part 10 · Up next

What comes next

Part 11 · Connections