Topic 6.3 Notes – Riemann Sums, Summation Notation, and Definite Integral Notation

What a Riemann Sum Is

A Riemann sum approximates the definite integral, which represents the signed area under a curve on an interval $[a,b]$.

You:

• Break $[a,b]$ into subintervals (a partition).
• Build rectangles on each subinterval.
• Add their areas.

Each rectangle has:

• Width $\Delta x_i$
• Height $f(x_i^*)$ where $x_i^*$ is some chosen point in that subinterval (left, right, or anywhere).

The general form is:

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

That’s literally “add up height × width.”

If the partition is equal:

• $\Delta x = \dfrac{b-a}{n}$
• A common sample point formula is $x_i = a + i\Delta x$

Here’s the geometric picture. Focus on how the rectangles use either the left endpoints or the right endpoints to determine height:

Left and right Riemann sums for a decreasing function

Because the function shown is decreasing, the left-endpoint sum overestimates the area and the right-endpoint sum underestimates it.

With a finite number of rectangles, this is an approximation.

Writing Riemann Sums in Sigma Notation

Sigma notation just makes the repeated addition compact.

Suppose you’re told:

• Interval $[a,b]$
• $n$ equal subintervals
• Right endpoints

You build it systematically:

1. Compute width

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

2. Write the sample point

   $$x_i = a + i\Delta x$$

3. Plug into the function and multiply by width:

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

Left endpoints usually use:

$$\sum_{i=0}^{n-1} f(a+i\Delta x)\,\Delta x$$

Small detail that matters on tests:
Right sums often run $i=1$ to $n$. Left sums often run $i=0$ to $n-1$. Don’t mix those.

On quizzes, you’ll either:

• Be asked to write the sum from a function and interval, or
• Be given a sigma expression and asked what integral it represents.

You need to move both directions comfortably.

The Definite Integral as a Limit

Now the key idea.

As the number of rectangles increases, widths shrink and the approximation improves.

When the widths approach 0:

$$\int_a^b f(x)\,dx = \lim_{\max \Delta x_i \to 0} \sum_{i=1}^{n} f(x_i^*)\,\Delta x_i$$

For equal partitions, this becomes:

$$\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(a+i\Delta x)\,\Delta x$$

where $\Delta x = \dfrac{b-a}{n}$.

Geometrically, you can picture the rectangles getting thinner and more numerous, so the jagged approximation hugs the curve more closely:

Riemann sums converging to a definite integral

Finite $n$ gives an approximation.
Letting $n \to \infty$ gives the exact signed area.

For AP purposes, this works cleanly when $f$ is continuous on $[a,b]$.

Converting Between Riemann Sums and Definite Integrals

Integral → Limit of a Riemann Sum

Given

$$\int_a^b f(x)\,dx$$

Rewrite it as:

$$\lim_{n\to\infty} \sum_{i=1}^{n} f\!\left(a+\frac{b-a}{n}i\right) \frac{b-a}{n}$$

Everything must be written in terms of $n$. That’s where students slip.

Limit of a Riemann Sum → Integral

If you see something like:

$$\lim_{n\to\infty} \sum_{i=1}^{n} g\!\left(2+\frac{5}{n}i\right) \frac{5}{n}$$

Notice:

• $\Delta x = \frac{5}{n} \Rightarrow b-a=5$
• Lower bound is 2
• So upper bound is $2+5=7$

That equals:

$$\int_2^7 g(x)\,dx$$

AP multiple choice loves disguising the interval inside that expression. Always extract:

• Lower bound $a$
• Width formula
• Upper bound $b=a+(b-a)$