Integration as the limit of a sum (AQA A-Level Mathematics): Revision Notes

1. Divide the Interval: Divide the interval $\ [a, b]$ into $\ n$ subintervals of equal width: $\\ \Delta x = \frac{b - a}{n}$

where $\ \Delta x$ is the width of each subinterval. 2. Choose Sample Points: In each subinterval, choose a sample point $\ x_i^*$ (this can be the left endpoint, right endpoint, midpoint, or any point within the subinterval). 3. Evaluate the Function: Compute the function value at each sample point $\ f(x_i^*) .$ 4. Form the Sum: Multiply the function value by the width of the subinterval $\ \Delta x$ to get the area of each rectangle. The Riemann sum is then the sum of these areas: $\\ S_n = \sum_{i=1}^n f(x_i^*) \Delta x$

Let's consider the function $\ f(x) = x^2$ over the interval $\ [0, 1]$.

Step 1: Set Up the Riemann Sum

Divide $\ [0, 1]$ into $\ n$ subintervals, each of width $\ \Delta x = \frac{1 - 0}{n} = \frac{1}{n} .$

For a right Riemann sum, the sample points are $\ x_i^* = \frac{i}{n} \ for \ i = 1, 2, \dots, n .$

Step 2: Compute the Riemann Sum

The Riemann sum is: $\\ S_n = \sum_{i=1}^n \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n}$

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

Step 3: Evaluate the Sum

The sum of squares of the first $\ n$ natural numbers is given by: $\\ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

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

Step 4: Take the Limit

Now, take the limit as $\ n \to \infty :$ $\\ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2}$

Simplify the expression: $\\ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2} = \frac{2}{6} = \frac{1}{3}$

So, the definite integral of $\ f(x) = x^2$ from $0$ to $1$ is$\ \frac{1}{3}$, which matches the result you would obtain using the Fundamental Theorem of Calculus: $\\ \int_0^1 x^2 \, dx = \frac{x^3}{3} \Big|_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$

• Integration as the limit of a sum is the foundational concept behind definite integration, where the integral is defined as the limit of a Riemann sum as the number of subintervals approaches infinity.
• Riemann sums approximate the area under a curve by summing up the areas of rectangles, and the exact area (the integral) is found by taking the limit of these sums.
• This approach underpins the definition of the definite integral and is central to understanding the relationship between area under a curve and integration.