Riemann Sums and the Definite Integral

By the end of this lesson, you will be able to set up and evaluate Riemann sums, explain how the definite integral is defined as a limit of Riemann sums, and interpret the definite integral as signed area.

The Area Problem

Given a non-negative function $f$ on $[a, b]$, we want to find the area between the curve and the x-axis. We can't use geometry directly (unless it's a simple shape). Instead: approximate with rectangles and take the limit.

Riemann Sums

Partition: divide $[a, b]$ into $n$ subintervals of equal width $\Delta x = \frac{b-a}{n}$.

Subintervals: $[x_0, x_1], [x_1, x_2], \ldots, [x_{n-1}, x_n]$ where $x_k = a + k\Delta x$.

A Riemann sum approximates the area using one rectangle per subinterval:

$$\sum_{k=1}^n f(x_k^*) \Delta x$$

where $x_k^*$ is a sample point chosen from the kth subinterval.

Types of Riemann sums:

• Left Riemann sum: $x_k^* = x_{k-1}$ (left endpoint)
• Right Riemann sum: $x_k^* = x_k$ (right endpoint)
• Midpoint Riemann sum: $x_k^* = \frac{x_{k-1}+x_k}{2}$ (midpoint)

For increasing functions: Left sum underestimates; Right sum overestimates. For decreasing functions: the opposite.

The Definite Integral as a Limit

As we use more and more rectangles (n → ∞), the Riemann sum approaches the exact area. This limit is the definite integral:

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

(This limit gives the same value regardless of which sample points $x_k^*$ are chosen, as long as $f$ is integrable.)

Computing a Riemann Sum Example

Approximate $\int_0^2 x^2, dx$ using a Right Riemann Sum with $n = 4$.

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

$x_1 = 0.5$, $x_2 = 1.0$, $x_3 = 1.5$, $x_4 = 2.0$

Right sum = $\sum_{k=1}^4 f(x_k) \cdot 0.5$

$= [f(0.5) + f(1.0) + f(1.5) + f(2.0)] \cdot 0.5$

$= [0.25 + 1 + 2.25 + 4] \cdot 0.5 = 7.5 \cdot 0.5 = 3.75$

Exact value: $\int_0^2 x^2, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} \approx 2.667$

The right sum overestimates because $f(x) = x^2$ is increasing on $[0, 2]$.