Area Approximation

Area approximation uses rectangles to estimate the area under a curve. Partition the interval into equal subintervals, build rectangles using left endpoints, right endpoints, or midpoints to determine height, then sum their areas.

Area Approximation Using Rectangles

When we need to find the area under a curve $f(x) \geq 0$ on an interval $[a,b]$ , we can approximate it by dividing the region into rectangles. This approach works even when we lack a closed-form antiderivative.

Setting Up the Partition

We partition $[a,b]$ into $n$ equal subintervals, each with width: $\Delta x = \frac{b-a}{n}$

The partition points are $x_0 = a, x_1, x_2, \ldots, x_n = b$ , where $x_i = a + i\Delta x$ . On each subinterval $[x_{i-1}, x_i]$ , we construct a rectangle of width $\Delta x$ . The height depends on where we sample the function value.

The Three Riemann Sums

For each subinterval, we can choose the function value at three different points:

Method	Formula
Left sum	$L_n = \sum_{i=1}^{n} f(x_{i-1})\,\Delta x$
Right sum	$R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x$
Midpoint sum	$M_n = \sum_{i=1}^{n} f\!\left(\frac{x_{i-1}+x_i}{2}\right)\Delta x$

Each sum represents the total area of all rectangles. As $n \to \infty$ , all three approach the true area.

Behavior Based on Function Monotonicity

For an increasing function, left endpoints give the smallest heights on each subinterval, making $L_n$ an underestimate. Right endpoints give the largest, making $R_n$ an overestimate. For decreasing functions, this reverses. The midpoint sum typically provides a better approximation than either endpoint method.

Example

For $f(x) = x^2 + 1$ on $[0,2]$ with $n=4$ subintervals ( $\Delta x = 0.5$ ):

$L_4 = 0.5[f(0)+f(0.5)+f(1)+f(1.5)] = 0.5(7.5) = 3.75$ $M_4 = 0.5[f(0.25)+f(0.75)+f(1.25)+f(1.75)] = 0.5(9.25) = 4.625$ $R_4 = 0.5[f(0.5)+f(1)+f(1.5)+f(2)] = 0.5(11.5) = 5.75$

Since $f$ is increasing, $L_4 < M_4 < R_4$ as expected.