MA 16500: Analytic Geometry and Calculus I

5.1. Approximating Area Under a Curve (Riemann Sums)

Goal: Understand how to approximate the area under a curve using Riemann sums and how this leads to the exact area via a limit.

1. Introduction to Riemann Sums

Suppose \( f(x) \) is continuous and \( f(x) \geq 0 \) on the interval \( [a,b] \). We want to compute the area between the graph of \( y = f(x) \) and the \( x \)-axis.

We approximate this area using the following steps:

  • Divide the interval \( [a,b] \) into \( n \) equal subintervals, each of width \[ \Delta x = \frac{b-a}{n}. \]
  • Choose a sample point \( x_k^* \) in each subinterval.
  • Construct rectangles with height \( f(x_k^*) \) and width \( \Delta x \).
  • The area of the \( k \)-th rectangle is \[ f(x_k^*) \cdot \Delta x. \]
  • Adding all rectangles gives the Riemann sum:
    \[ A_n = \sum_{k=1}^{n} f(x_k^*) \, \Delta x. \]
Three graphs comparing Left, Right, and Midpoint Riemann sums using rectangles to approximate the area under a curve.
Three graphs comparing Left, Right, and Midpoint Riemann sums using rectangles to approximate the area under a curve.

Definition: If \( f(x) \) is continuous on \( [a,b] \), then the exact area under the curve is defined as the limit of the Riemann sums:

\[ A = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k^*) \, \Delta x. \]

This limit exists when \( f(x) \) is continuous, and it is used as the definition of the true area under the curve.

2. Examples

Example 2.1: Area Using Riemann Sum

We consider the area enclosed by the line \( y = 2x \), the line \( x = 3 \), and the \( x \)-axis.

Of course, this area can be computed geometrically as the area of a triangle:

\[ A = \frac{1}{2} \cdot 3 \cdot 6 = 9. \]

We now compute the same area using a Riemann sum.

Solution:

Divide the interval \( [0,3] \) into \( n \) equal subintervals. Then

\[ \Delta x = \frac{3}{n}. \]

Using right endpoints, the sample points are

\[ x_k = \frac{3k}{n}, \quad k = 1,2,\dots,n. \]

The function values are

\[ f(x_k) = 2x_k = 2\left(\frac{3k}{n}\right) = \frac{6k}{n}. \]

The Riemann sum is

\[ A_n = \sum_{k=1}^{n} f(x_k)\,\Delta x = \sum_{k=1}^{n} \frac{6k}{n} \cdot \frac{3}{n} = \frac{18}{n^2} \sum_{k=1}^{n} k. \]

Using the formula \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \), we get

\[ A_n = \frac{18}{n^2} \cdot \frac{n(n+1)}{2} = 9 \cdot \frac{n+1}{n}. \]

Taking the limit as \( n \to \infty \),

\[ \lim_{n \to \infty} A_n = 9. \]

This agrees with the geometric result, confirming that the exact area is \( A = 9 \).

Example 2.2: Area Under the Parabola \(y = x^2\)

We consider the area enclosed by the parabola \( y = x^2 \), the line \( x = 1 \), and the \( x \)-axis.

Geometrically, there is no simple formula like a triangle, so we compute the area using a Riemann sum.

Solution:

Divide the interval \( [0,1] \) into \( n \) equal subintervals:

\[ \Delta x = \frac{1-0}{n} = \frac{1}{n}. \]

Using right endpoints, the sample points are

\[ x_k = \frac{k}{n}, \quad k = 1,2,\dots,n. \]

Function values at these points are

\[ f(x_k) = x_k^2 = \left(\frac{k}{n}\right)^2 = \frac{k^2}{n^2}. \]

The Riemann sum is

\[ A_n = \sum_{k=1}^{n} f(x_k)\,\Delta x = \sum_{k=1}^{n} \frac{k^2}{n^2} \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{k=1}^{n} k^2. \]

Using the formula \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \), we get

\[ A_n = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}. \]

Taking the limit as \( n \to \infty \):

\[ \lim_{n \to \infty} A_n = \frac{1 \cdot 2}{6} = \frac{1}{3}. \]

Therefore, the exact area under the parabola from \( x = 0 \) to \( x = 1 \) is \( A = \frac{1}{3} \).