5.1 Approximating Areas

Sigma notation gives you a compact way to write long sums without listing every term. It's the language you'll use to express Riemann sums, which are the foundation for understanding how integration works.

Area estimation using rectangles is how we build toward the definite integral. By chopping a region into thin rectangles, summing their areas, and then taking a limit, you move from approximation to exact area.

Sigma Notation and Summations

Summations with sigma notation

Sigma notation uses the Greek letter $\Sigma$ to represent the sum of a series of terms. Instead of writing out $1^2 + 2^2 + 3^2 + \cdots + 10^2$, you write $\sum_{i=1}^{10} i^2$. Much cleaner.

Here's how to read it:

• The index of summation (usually $i$, $k$, or $n$) appears below the $\Sigma$ along with its starting value.
• The upper limit above the $\Sigma$ tells you where to stop.
• The general term to the right of $\Sigma$ is the expression you evaluate for each value of the index.

So $\sum_{i=1}^{n} a_i$ means "plug in $i = 1$, then $i = 2$, all the way up to $i = n$, and add up every $a_i$ you get."

Evaluating a summation step by step:

1. Identify the starting value, ending value, and general term.
2. Substitute each index value into the general term.
3. Add the results.

For example, $\sum_{i=1}^{4} (2i + 1) = (2(1)+1) + (2(2)+1) + (2(3)+1) + (2(4)+1) = 3 + 5 + 7 + 9 = 24$.

A few summation properties worth knowing (these save time on longer problems):

• $\sum_{i=1}^{n} c = cn$ (summing a constant $n$ times)
• $\sum_{i=1}^{n} c \cdot a_i = c \sum_{i=1}^{n} a_i$ (constants factor out)
• $\sum_{i=1}^{n} (a_i + b_i) = \sum_{i=1}^{n} a_i + \sum_{i=1}^{n} b_i$ (sums split apart)

You'll also use these closed-form formulas frequently:

• $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$
• $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$
• $\sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2$

These let you evaluate Riemann sums without plugging in every single term.

Approximating Areas

Area estimation using rectangles

The core idea: to find the area under a curve $y = f(x)$ from $x = a$ to $x = b$, you approximate it with rectangles whose areas are easy to compute.

Setting up the rectangles:

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

2. On each subinterval, build a rectangle with width $\Delta x$.

3. Choose a point in each subinterval to determine the rectangle's height. The three standard choices are:

   • Left endpoint: use $f(x_{i-1})$, the function value at the left edge of each subinterval
   • Right endpoint: use $f(x_i)$, the function value at the right edge
   • Midpoint: use $f\!\left(\frac{x_{i-1} + x_i}{2}\right)$, the function value at the center

4. Each rectangle has area $A_i = f(x_i^*) \cdot \Delta x$, where $x_i^*$ is your chosen sample point.

5. Sum all rectangle areas: $A \approx \sum_{i=1}^{n} f(x_i^*)\,\Delta x$.

Quick example: Approximate the area under $f(x) = x^2$ on $[0, 2]$ using $n = 4$ right-endpoint rectangles.

• $\Delta x = \frac{2 - 0}{4} = 0.5$
• Right endpoints: $x = 0.5, 1.0, 1.5, 2.0$
• Heights: $f(0.5) = 0.25$, $f(1) = 1$, $f(1.5) = 2.25$, $f(2) = 4$
• Area $\approx (0.25 + 1 + 2.25 + 4)(0.5) = 7.5(0.5) = 3.75$

The exact answer is $\frac{8}{3} \approx 2.667$, so this right-endpoint sum overestimates because $x^2$ is increasing on $[0, 2]$. That's a pattern: for an increasing function, right sums overestimate and left sums underestimate. For a decreasing function, it's the opposite.

Riemann sums for definite integrals

A Riemann sum is the formal name for the rectangle-area approximation described above. The general formula is:

$\sum_{i=1}^{n} f(x_i^*)\,\Delta x$

where $\Delta x = \frac{b-a}{n}$ and $x_i^*$ is a sample point in the $i$-th subinterval.

The type of Riemann sum depends on your choice of sample point:

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

This limit exists (and gives the same value regardless of sample point choice) whenever $f$ is continuous on $[a, b]$.

Approximation techniques and limits

The whole point of this section is the transition from approximation to exact value through limits.

A partition of $[a, b]$ is just the set of points that divide the interval into subintervals: $a = x_0 < x_1 < x_2 < \cdots < x_n = b$. With equal subintervals, each has width $\Delta x$, and the endpoints are $x_i = a + i\,\Delta x$.

Here's the process for finding an exact area using a limit of Riemann sums:

1. Write the Riemann sum in sigma notation using the formulas for $x_i$ and $\Delta x$.
2. Expand $f(x_i^*)$ and simplify using summation formulas (the closed-form formulas from the sigma notation section).
3. Take $\lim_{n \to \infty}$ of the resulting expression.

For instance, to find $\int_0^1 x^2\,dx$ using right endpoints:

1. $\Delta x = \frac{1}{n}$, and $x_i = \frac{i}{n}$
2. $\sum_{i=1}^{n} \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n} = \frac{1}{n^3}\sum_{i=1}^{n} i^2 = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}$
3. $\lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim_{n \to \infty} \frac{2n^2 + 3n + 1}{6n^2} = \frac{2}{6} = \frac{1}{3}$

So $\int_0^1 x^2\,dx = \frac{1}{3}$. This is exactly what the Fundamental Theorem of Calculus will confirm later, but here you've derived it from first principles using Riemann sums.