📄 Save or print this lesson:

← Module 6 Home Lesson 2 of 4 Next Lesson →

Lesson 2: Riemann Sums and Approximating Area

Estimated time: 40-50 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Use sigma notation to express sums compactly
Compute left, right, and midpoint Riemann sums
Determine whether a Riemann sum overestimates or underestimates the true area
Understand that the definite integral is the limit of Riemann sums as n → ∞

The Area Problem

One of the central problems in calculus is computing the area of a region bounded by a curve, the x-axis, and two vertical lines x = a and x = b. For simple shapes like rectangles and triangles we have formulas, but what about the area under y = x² from x = 0 to x = 1? We need a new strategy: approximate the region with rectangles, then take a limit.

Key Idea

Divide the interval [a, b] into n subintervals of equal width Δx = (b − a)/n. On each subinterval, build a rectangle whose height is determined by the function value at a chosen sample point. The sum of the rectangle areas approximates the area under the curve.

Sigma Notation Review

Sigma notation provides a compact way to write sums. The Greek letter Σ (sigma) means "add up":

Sigma Notation

∑_i=1ⁿ a_i = a₁ + a₂ + … + a_n

Here i is the index of summation, 1 is the lower limit, and n is the upper limit.

Useful formulas:

∑_i=1ⁿ c = cn (sum of a constant)
∑_i=1ⁿ i = n(n + 1)/2
∑_i=1ⁿ i² = n(n + 1)(2n + 1)/6
∑_i=1ⁿ i³ = [n(n + 1)/2]²

Example 1: Evaluating a Sigma Sum

Evaluate ∑_i=1⁴ (2i + 1).

Step 1: Write out each term: (2(1)+1) + (2(2)+1) + (2(3)+1) + (2(4)+1).

Step 2: Compute: 3 + 5 + 7 + 9 = 24.

Alternatively: ∑(2i + 1) = 2∑i + ∑1 = 2(4)(5)/2 + 4 = 20 + 4 = 24.

Left Riemann Sums

Left Riemann Sum (L_n)

Use the left endpoint of each subinterval as the sample point:

L_n = ∑_i=0ⁿ⁻¹ f(x_i) · Δx

where x_i = a + i · Δx.

Example 2: Left Riemann Sum

Approximate the area under f(x) = x² on [0, 2] using n = 4 subintervals with a left Riemann sum.

Step 1: Δx = (2 − 0)/4 = 0.5.

Step 2: Left endpoints: x₀ = 0, x₁ = 0.5, x₂ = 1, x₃ = 1.5.

Step 3: Function values: f(0) = 0, f(0.5) = 0.25, f(1) = 1, f(1.5) = 2.25.

Step 4: L₄ = (0 + 0.25 + 1 + 2.25)(0.5) = 3.5(0.5) = 1.75.

Since f(x) = x² is increasing on [0, 2], the left sum underestimates the true area.

Right Riemann Sums

Right Riemann Sum (R_n)

Use the right endpoint of each subinterval:

R_n = ∑_i=1ⁿ f(x_i) · Δx

where x_i = a + i · Δx.

Example 3: Right Riemann Sum

Approximate the area under f(x) = x² on [0, 2] using n = 4 with a right sum.

Step 1: Δx = 0.5 (same as before).

Step 2: Right endpoints: x₁ = 0.5, x₂ = 1, x₃ = 1.5, x₄ = 2.

Step 3: f(0.5) = 0.25, f(1) = 1, f(1.5) = 2.25, f(2) = 4.

Step 4: R₄ = (0.25 + 1 + 2.25 + 4)(0.5) = 7.5(0.5) = 3.75.

Since f is increasing, the right sum overestimates. The true area (8/3 ≈ 2.667) lies between L₄ = 1.75 and R₄ = 3.75.

Midpoint Riemann Sums

Midpoint Riemann Sum (M_n)

Use the midpoint of each subinterval:

M_n = ∑_i=1ⁿ f(x̄_i) · Δx

where x̄_i = (x_i−1 + x_i)/2 is the midpoint of the i-th subinterval.

Example 4: Midpoint Sum

Approximate the area under f(x) = x² on [0, 2] with n = 4 using midpoints.

Step 1: Δx = 0.5. Midpoints: 0.25, 0.75, 1.25, 1.75.

Step 2: f(0.25) = 0.0625, f(0.75) = 0.5625, f(1.25) = 1.5625, f(1.75) = 3.0625.

Step 3: M₄ = (0.0625 + 0.5625 + 1.5625 + 3.0625)(0.5) = 5.25(0.5) = 2.625.

This is much closer to the true value 8/3 ≈ 2.667 than either L₄ or R₄. Midpoint sums typically give better approximations.

Over- and Underestimates

Whether a sum overestimates or underestimates depends on the function's behavior:

Increasing function: Left sum underestimates; right sum overestimates.
Decreasing function: Left sum overestimates; right sum underestimates.
Concave up: Midpoint sum underestimates; trapezoidal sum overestimates.
Concave down: Midpoint sum overestimates; trapezoidal sum underestimates.

As n increases (more, thinner rectangles), all Riemann sums converge to the same value: the definite integral.

Example 5: Computing with Sigma Formulas

Use the formula for ∑i² to find the exact area under f(x) = x² on [0, 2] via a right Riemann sum as n → ∞.

Step 1: Δx = 2/n, x_i = 2i/n.

Step 2: R_n = ∑_i=1ⁿ (2i/n)² · (2/n) = ∑_i=1ⁿ (4i²/n²)(2/n) = (8/n³) ∑_i=1ⁿ i².

Step 3: Substitute the formula: = (8/n³) · n(n+1)(2n+1)/6 = 8(n+1)(2n+1)/(6n²).

Step 4: Take the limit: lim_n→∞ 8(n+1)(2n+1)/(6n²) = 8(2)/6 = 8/3.

Example 6: Riemann Sum from a Table

A car's velocity v(t) in ft/s is measured every 2 seconds:

t	0	2	4	6
v(t)	10	18	24	28

Estimate the total distance traveled on [0, 6] using a left Riemann sum.

Solution: Δt = 2. L₃ = v(0)·2 + v(2)·2 + v(4)·2 = 10(2) + 18(2) + 24(2) = 20 + 36 + 48 = 104 ft.

Key Takeaways

Riemann sums approximate area by summing rectangle areas: ∑ f(x_i*) · Δx.
Left, right, and midpoint sums differ only in where the sample point is chosen in each subinterval.
For monotone functions, left and right sums bound the true area from below and above.
The midpoint sum usually gives a better approximation than left or right sums for the same n.
The exact area equals lim_n→∞ ∑ f(x_i*) Δx, which is the definite integral.

Check Your Understanding

1. Compute the right Riemann sum for f(x) = 3x on [1, 5] with n = 4.

Answer: Δx = 1. Right endpoints: 2, 3, 4, 5. R₄ = [3(2) + 3(3) + 3(4) + 3(5)](1) = 6 + 9 + 12 + 15 = 42. (The true area is 36.)

2. If f is decreasing on [a, b], does the left Riemann sum overestimate or underestimate?

Answer: Overestimates. For a decreasing function, the left endpoint gives the largest value on each subinterval, so each rectangle is taller than the curve.

3. Evaluate ∑_i=1⁵ (i² + 1).

Answer: ∑i² + ∑1 = 5(6)(11)/6 + 5 = 55 + 5 = 60.

4. Using n = 3 subintervals and a midpoint sum, approximate the area under f(x) = 1/x on [1, 4].

Answer: Δx = 1. Midpoints: 1.5, 2.5, 3.5. M₃ = [1/1.5 + 1/2.5 + 1/3.5](1) = 0.6667 + 0.4 + 0.2857 ≈ 1.352. (True value: ln 4 ≈ 1.386.)

Ready for More?

Next Lesson

In Lesson 3, you will define the definite integral as a limit of Riemann sums and discover the Fundamental Theorem of Calculus Part 1.

Start Lesson 3

Module Progress

You have completed Lesson 2! Keep going to master integration.

Back to Module Home