Approximating Areas with Riemann Sums

3. Visual Connection

5. Concept Development

Approximation works across representations

A definite integral can be approximated when a function is given by a formula, table, graph, or verbal description. The central idea is always the same: divide the interval into subintervals, choose a representative height on each one, multiply height by width, and add the signed contributions.

Build the partition first

A partition \(a=x_0<x_1<\cdots<x_n=b\) creates the subintervals \([x_{i-1},x_i]\). Write these intervals before choosing heights; this prevents endpoint shifts and missing terms.

Uniform partitions

For \(n\) equal-width subintervals of \([a,b]\), every rectangle has width

The number \(n\) counts subintervals, so a partition with \(n\) intervals has \(n+1\) endpoints.

Nonuniform partitions

When table inputs are unevenly spaced, use the individual widths \(\Delta x_i=x_i-x_{i-1}\). There is no common \(\Delta x\) to factor out:

Left-endpoint sums

The left sum uses the first endpoint of each subinterval:

The final table value is not used as a height because it is not the left endpoint of any included subinterval.

Right-endpoint sums

The right sum uses the second endpoint of each subinterval:

The initial table value is not used as a height. Each right endpoint must be paired with the width immediately to its left.

Midpoint sums

The midpoint sum samples each interval at \(m_i=(x_{i-1}+x_i)/2\):

A table supports this method only if it supplies, or allows calculation of, the function values at those midpoints.

Trapezoidal sums

A trapezoid uses the average of its two endpoint heights:

This formula works for both uniform and nonuniform partitions.

Trapezoids average left and right sums

For the same partition, each trapezoid averages its left and right rectangle. Consequently,

This identity is a useful calculation check, but it does not make the trapezoidal estimate automatically exact.

Keep signed heights

If \(f(x_i^*)<0\), the corresponding rectangle contributes a negative amount. A Riemann sum approximates a definite integral, or net signed area. If geometric area is requested, split at zeros and make below-axis contributions positive.

Monotonicity controls left and right error

For an increasing function, left rectangles lie below the curve and right rectangles lie above it, so \(L\) underestimates and \(R\) overestimates. For a decreasing function, the conclusions reverse. Without known monotonicity, no general left/right error claim is justified.

Concavity controls midpoint and trapezoidal error

For a concave-up function, secant segments lie above the graph, so the trapezoidal sum overestimates; midpoint rectangles underestimate. For a concave-down function, trapezoidal sums underestimate and midpoint sums overestimate.

More subintervals and accuracy

For a well-behaved function, making the largest subinterval width smaller generally improves the approximation and the sums approach one limiting value. However, a single increase in \(n\) does not guarantee that every named approximation is closer, so compare error using supported behavior rather than intuition alone.

Interpret estimates in context

If the integrand is a rate, the product of rate and input width has units of accumulated quantity. State that the answer is an approximation, include units, and distinguish estimated change from a final amount.

Common errors

Frequent errors include using \(n\) data points as though they made \(n\) intervals, pairing a height with the wrong width, assuming table intervals are equal, using unavailable midpoint values, dropping negative signs, and deciding over- or underestimation from concavity when the method is left or right.

6. Detailed Worked Example and Error Check

Example 1: Four methods on one function. Approximate \(\int_0^2(x^2+1)\,dx\) with four equal subintervals. Here \(\Delta x=1/2\).

At the midpoints \(0.25,0.75,1.25,1.75\), the heights sum to \(9.25\), so \(M_4=4.625\). Also, \(T_4=(L_4+R_4)/2=4.75\). The exact value is \(14/3\approx4.667\): increasing behavior explains the left/right errors, and concavity up explains the midpoint/trapezoidal errors.

Example 2: Nonuniform table. Suppose

The widths are \(1,2,3\), so

The trapezoidal estimate is \(T=(24+43)/2=33.5\). A midpoint estimate cannot be computed from this table because the required midpoint values are missing.

Example 3: A rate table and units. A flow rate \(r(t)\), in liters per hour, is recorded at times \(0,2,5,9\) hours with values \(3,6,4,-2\). A trapezoidal estimate of net change is

The negative endpoint remains part of the signed estimate, and the unequal time widths are handled separately.

Example 4: Reading monotonicity. A positive function decreases on \([1,5]\). On every subinterval, the left endpoint gives the larger height and the right endpoint gives the smaller height. Therefore

No concavity information is needed for this left/right comparison.

Example 5: Reading concavity. Suppose \(g''(x)<0\) on \([0,6]\). The graph is concave down, so midpoint rectangles tend to sit above the curve while secant-topped trapezoids lie below it:

Whether \(g\) increases or decreases is irrelevant to this midpoint/trapezoidal error direction.

Example 6: Linear functions. For \(h(x)=3x-2\), the graph itself is the secant line on every subinterval. Thus every trapezoidal estimate is exact, regardless of the partition. Midpoint estimates are also exact because a linear function's midpoint height equals the average of its endpoint heights.

Example 7: Signed table estimate. A velocity table on equal two-second intervals gives right-endpoint values \(5,-1,-4,3\) meters per second. The right Riemann sum is

This estimates displacement. Replacing the values by their absolute values would instead estimate total distance: \(2(5+1+4+3)=26\) meters.