For simple shapes (rectangles, triangles), we have formulas. But for curves? We need a new approach!

💡 Key Idea: Approximate the area using rectangles, then take the limit as the number of rectangles approaches infinity!

Approximating with Rectangles

The Basic Strategy

Step 1: Divide the interval $[a, b]$ into $n$ subintervals

Step 2: On each subinterval, draw a rectangle

Step 3: Add up the areas of all rectangles

Step 4: Take the limit as $n \to \infty$

Partitioning the Interval

Regular Partition

Divide $[a, b]$ into $n$ equal subintervals:

Width of each rectangle: $\Delta x = \frac{b-a}{n}$

Partition points: $x_0 = a, \quad x_1 = a + \Delta x, \quad x_2 = a + 2\Delta x, \quad \ldots, \quad x_n = b$

In general: $x_i = a + i\Delta x$

Types of Riemann Sums

The height of each rectangle depends on where we sample the function!

Left Riemann Sum

Use the left endpoint of each subinterval:

$L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x = f(x_0)\Delta x + f(x_1)\Delta x + \cdots + f(x_{n-1})\Delta x$

Right Riemann Sum

Use the right endpoint of each subinterval:

$R_n = \sum_{i=1}^{n} f(x_i) \Delta x = f(x_1)\Delta x + f(x_2)\Delta x + \cdots + f(x_n)\Delta x$

Midpoint Riemann Sum

Use the midpoint of each subinterval:

$M_n = \sum_{i=1}^{n} f\left(\frac{x_{i-1} + x_i}{2}\right) \Delta x$

Example 1: Computing a Left Riemann Sum

Approximate the area under $f(x) = x^2$ from $x = 0$ to $x = 2$ using 4 rectangles (left endpoints).

Step 1: Find $\Delta x$

$\Delta x = \frac{b-a}{n} = \frac{2-0}{4} = \frac{1}{2}$

Step 2: Find partition points

$x_0 = 0$

$x_1 = 0 + \frac{1}{2} = 0.5$

$x_2 = 0.5 + \frac{1}{2} = 1$

$x_3 = 1 + \frac{1}{2} = 1.5$

$x_4 = 1.5 + \frac{1}{2} = 2$

Step 3: Calculate function values at left endpoints

$f(x_0) = f(0) = 0^2 = 0$

$f(x_1) = f(0.5) = (0.5)^2 = 0.25$

$f(x_2) = f(1) = 1^2 = 1$

$f(x_3) = f(1.5) = (1.5)^2 = 2.25$

Step 4: Calculate Left Riemann Sum

$L_4 = [f(x_0) + f(x_1) + f(x_2) + f(x_3)] \cdot \Delta x$

$= [0 + 0.25 + 1 + 2.25] \cdot \frac{1}{2}$

$= 3.5 \cdot 0.5 = 1.75$

Answer: The left Riemann sum approximation is 1.75 square units.

Note: The actual area is $\frac{8}{3} \approx 2.67$, so this is an underestimate (because $f$ is increasing).

Example 2: Right Riemann Sum

Same function $f(x) = x^2$ from $x = 0$ to $x = 2$ with 4 rectangles (right endpoints).

Use the same partition, but sample at right endpoints:

$f(x_1) = f(0.5) = 0.25$

$f(x_2) = f(1) = 1$

$f(x_3) = f(1.5) = 2.25$

$f(x_4) = f(2) = 4$

$R_4 = [f(x_1) + f(x_2) + f(x_3) + f(x_4)] \cdot \Delta x$

$= [0.25 + 1 + 2.25 + 4] \cdot \frac{1}{2}$

$= 7.5 \cdot 0.5 = 3.75$

Answer: The right Riemann sum approximation is 3.75 square units.

This is an overestimate (because $f$ is increasing and we use right endpoints).

Increasing vs Decreasing Functions

For Increasing Functions

• Left sum underestimates (rectangles below curve)
• Right sum overestimates (rectangles above curve)
• Midpoint sum is usually more accurate

For Decreasing Functions

• Left sum overestimates
• Right sum underestimates
• Midpoint sum is usually more accurate

Sigma Notation Review

Summation Symbol

$\sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \cdots + a_n$

Read as: "The sum from $i = 1$ to $n$ of $a_i$"

Useful Formulas

$\sum_{i=1}^{n} c = cn$ (sum of constants)

$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ (sum of first $n$ integers)

$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ (sum of squares)

$\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2$ (sum of cubes)

Example 3: Using Formulas

Find the right Riemann sum for $f(x) = x$ on $[0, 1]$ with $n$ rectangles, then take the limit.

Step 1: Find $\Delta x$

$\Delta x = \frac{1-0}{n} = \frac{1}{n}$

Step 2: Partition points

$x_i = 0 + i\Delta x = \frac{i}{n}$

Step 3: Right Riemann sum

$R_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} f\left(\frac{i}{n}\right) \cdot \frac{1}{n}$

$= \sum_{i=1}^{n} \frac{i}{n} \cdot \frac{1}{n} = \sum_{i=1}^{n} \frac{i}{n^2}$

$= \frac{1}{n^2} \sum_{i=1}^{n} i = \frac{1}{n^2} \cdot \frac{n(n+1)}{2}$

$= \frac{n+1}{2n} = \frac{1}{2} + \frac{1}{2n}$

Step 4: Take the limit

$\lim_{n \to \infty} R_n = \lim_{n \to \infty} \left(\frac{1}{2} + \frac{1}{2n}\right) = \frac{1}{2}$

Answer: The exact area is $\frac{1}{2}$ square unit.

Check: Area of triangle with base 1 and height 1 is $\frac{1}{2}(1)(1) = \frac{1}{2}$ ✓

The Definite Integral (Preview)

As $n \to \infty$, the Riemann sum approaches the definite integral:

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

This is the exact area under the curve!

Properties of Riemann Sums

More Rectangles = Better Approximation

As $n$ increases (more rectangles, thinner width):

• The approximation gets more accurate
• Left, right, and midpoint sums all approach the same value

Convergence

For continuous functions on $[a, b]$: $\lim_{n \to \infty} L_n = \lim_{n \to \infty} R_n = \lim_{n \to \infty} M_n = \int_a^b f(x)\,dx$

Trapezoidal Rule

Instead of rectangles, use trapezoids!

$T_n = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)]$

Pattern: First and last get coefficient 1, all middle terms get coefficient 2.

Often more accurate than left/right/midpoint sums!

Example 4: Trapezoidal Rule

Approximate $\int_0^2 x^2\,dx$ using 4 trapezoids.

Step 1: $\Delta x = \frac{2-0}{4} = 0.5$

Points: $x_0=0, x_1=0.5, x_2=1, x_3=1.5, x_4=2$

Function values: $f(x_0)=0, f(x_1)=0.25, f(x_2)=1, f(x_3)=2.25, f(x_4)=4$

Step 2: Apply formula

$T_4 = \frac{0.5}{2}[0 + 2(0.25) + 2(1) + 2(2.25) + 4]$

$= 0.25[0 + 0.5 + 2 + 4.5 + 4]$

$= 0.25 \cdot 11 = 2.75$

Actual: $\frac{8}{3} \approx 2.667$

Pretty close! Better than left (1.75) or right (3.75) with same number of rectangles.

⚠️ Common Mistakes

Mistake 1: Wrong Number of Terms

For $n$ rectangles:

• Left sum uses $x_0, x_1, \ldots, x_{n-1}$ ($n$ terms)
• Right sum uses $x_1, x_2, \ldots, x_n$ ($$ terms)

Don't mix them up!

Mistake 2: Forgetting $\Delta x$

Each rectangle has area = height × width!

Don't forget to multiply by $\Delta x$.

Mistake 3: Using Wrong Endpoints

Left sum: use left endpoints ($x_0$ through $x_{n-1}$)

Right sum: use right endpoints ($x_1$ through $x_n$)

Mistake 4: Arithmetic Errors

With multiple terms, it's easy to make calculation mistakes.

Check: Does your answer make sense? Is it positive when area should be positive?

Historical Note

Named after Bernhard Riemann (1826-1866), who formalized the concept of integration using these sums.

The idea dates back to Archimedes (~250 BC) who used "method of exhaustion" to find areas!

The Big Picture

Riemann sums are the foundation of the definite integral:

1. Approximate area with rectangles
2. More rectangles = better approximation
3. Take the limit as $n \to \infty$
4. Get exact area = definite integral

Next, we'll learn the Fundamental Theorem of Calculus, which gives us an easier way to compute these areas!

📝 Practice Strategy

1. Find $\Delta x = \frac{b-a}{n}$ first
2. List partition points: $x_i = a + i\Delta x$
3. Identify which endpoints to use (left, right, or midpoint)
4. Calculate function values at those points
5. Sum and multiply by $\Delta x$
6. Check: Does the answer make sense given the graph?
7. For limits: Use summation formulas, then let $n \to \infty$

Step 2: Find partition points

$x_0 = 0, \quad x_1 = 1, \quad x_2 = 2, \quad x_3 = 3$

Step 3: Find midpoints

Midpoint of $[x_0, x_1]$: $m_1 = \frac{0+1}{2} = 0.5$

Midpoint of $[x_1, x_2]$: $m_2 = \frac{1+2}{2} = 1.5$

Midpoint of $[x_2, x_3]$: $m_3 = \frac{2+3}{2} = 2.5$

Step 4: Calculate function values at midpoints

$f(x) = x^2$

$f(m_1) = f(0.5) = (0.5)^2 = 0.25$

$f(m_2) = f(1.5) = (1.5)^2 = 2.25$

$f(m_3) = f(2.5) = (2.5)^2 = 6.25$

Step 5: Calculate Midpoint Riemann Sum

$M_3 = [f(m_1) + f(m_2) + f(m_3)] \cdot \Delta x$

$= [0.25 + 2.25 + 6.25] \cdot 1$

$= 8.75$

Answer: The midpoint Riemann sum approximation is 8.75 square units.

Note: The actual value is $\int_0^3 x^2\,dx = \frac{x^3}{3}\Big|_0^3 = 9$, so this is quite accurate!

$x_i = 0 + i \cdot \frac{1}{n} = \frac{i}{n}$

Step 2: Write the sum

$R_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} f\left(\frac{i}{n}\right) \cdot \frac{1}{n}$

$= \sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n}$

$= \sum_{i=1}^{n} \frac{i^2}{n^2} \cdot \frac{1}{n} = \sum_{i=1}^{n} \frac{i^2}{n^3}$

Step 3: Factor out constant

$= \frac{1}{n^3} \sum_{i=1}^{n} i^2$

Step 4: Use summation formula

$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

$R_n = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$

$= \frac{n(n+1)(2n+1)}{6n^3}$

Step 5: Simplify

$= \frac{(n+1)(2n+1)}{6n^2} = \frac{2n^2 + 3n + 1}{6n^2}$

$= \frac{2n^2}{6n^2} + \frac{3n}{6n^2} + \frac{1}{6n^2}$

$= \frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}$

Step 6: Take the limit

$\lim_{n \to \infty} R_n = \lim_{n \to \infty} \left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right)$

$= \frac{1}{3} + 0 + 0 = \frac{1}{3}$

Answer: $\int_0^1 x^2\,dx = \frac{1}{3}$