The Central Idea
Suppose we want to estimate the area beneath a curve
\[ y=f(x) \]from
\[ x=a \]to
\[ x=b. \]Divide the interval into smaller subintervals. On each subinterval, choose a representative function value and use it as the height of a rectangle.
Add the rectangle areas:
\[ \boxed{ \text{estimated area} = \sum (\text{rectangle height})(\text{rectangle width}) }. \]Why Rectangles?
Rectangles are useful because their areas are easy to compute:
\[ A=bh. \]If the interval is narrow enough, the graph of a smooth function changes only a small amount across that interval.
A rectangle can then serve as a simple approximation to the curved strip.
Many small rectangles can approximate a curved region far better than one large rectangle.
Partitioning an Interval
A partition divides an interval into subintervals.
Suppose
\[ [a,b] \]is divided into \(n\) equal pieces.
The width of each piece is
\[ \boxed{ \Delta x=\frac{b-a}{n} }. \]This quantity becomes the width of each approximating rectangle.
Example: Find the Rectangle Width
Divide
\[ [0,8] \]into \(4\) equal subintervals.
\[ \Delta x = \frac{8-0}{4} = \boxed{2}. \]The partition points are
\[ 0,\ 2,\ 4,\ 6,\ 8. \]Rectangle Heights
Once the interval is divided, we must decide where to measure the function value used as the rectangle height.
Common choices are:
- the left endpoint of each subinterval;
- the right endpoint of each subinterval;
- the midpoint of each subinterval.
These choices produce different numerical approximations.
Left-Endpoint Rectangles
In a left-endpoint approximation, the rectangle height on each subinterval is determined by the function value at the left endpoint.
If the partition points are
\[ x_0,x_1,\ldots,x_n, \]then the left-endpoint heights are
\[ f(x_0),\ f(x_1),\ldots,f(x_{n-1}). \]For equal widths,
\[ \boxed{ L_n = \Delta x \left[ f(x_0)+f(x_1)+\cdots+f(x_{n-1}) \right] }. \]Right-Endpoint Rectangles
In a right-endpoint approximation, each height comes from the right endpoint of the subinterval.
The heights are
\[ f(x_1),\ f(x_2),\ldots,f(x_n). \]Therefore,
\[ \boxed{ R_n = \Delta x \left[ f(x_1)+f(x_2)+\cdots+f(x_n) \right] }. \]Midpoint Rectangles
In a midpoint approximation, each rectangle height is taken from the midpoint of its subinterval.
If the midpoint of the \(i\)th subinterval is \(m_i\), then
\[ \boxed{ M_n = \Delta x \left[ f(m_1)+f(m_2)+\cdots+f(m_n) \right] }. \]Midpoint rectangles often balance error more effectively than using only one side of each interval.
Example Function
Consider
\[ f(x)=x^2 \]on
\[ [0,2]. \]Divide the interval into \(4\) equal subintervals.
\[ \Delta x = \frac{2-0}{4} = \boxed{0.5}. \]The partition points are
\[ 0,\ 0.5,\ 1,\ 1.5,\ 2. \]Left Approximation for \(x^2\)
The left endpoints are
\[ 0,\ 0.5,\ 1,\ 1.5. \]Evaluate:
\[ f(0)=0, \] \[ f(0.5)=0.25, \] \[ f(1)=1, \] \[ f(1.5)=2.25. \]Therefore,
\[ L_4 = 0.5(0+0.25+1+2.25). \] \[ L_4 = 0.5(3.5) = \boxed{1.75}. \]Right Approximation for \(x^2\)
The right endpoints are
\[ 0.5,\ 1,\ 1.5,\ 2. \]Evaluate:
\[ 0.25,\ 1,\ 2.25,\ 4. \]Therefore,
\[ R_4 = 0.5(0.25+1+2.25+4). \] \[ R_4 = 0.5(7.5) = \boxed{3.75}. \]Midpoint Approximation for \(x^2\)
The midpoints are
\[ 0.25,\ 0.75,\ 1.25,\ 1.75. \]Their squared values are
\[ 0.0625,\ 0.5625,\ 1.5625,\ 3.0625. \]Therefore,
\[ M_4 = 0.5(0.0625+0.5625+1.5625+3.0625). \] \[ M_4 = 0.5(5.25) = \boxed{2.625}. \]Comparing the Three Estimates
For \(f(x)=x^2\) on \([0,2]\),
\[ L_4=1.75, \] \[ R_4=3.75, \]and
\[ M_4=2.625. \]The exact area, developed later through integration, is
\[ \frac{8}{3}\approx2.667. \]The midpoint estimate is especially close in this example.
Increasing Functions and Left Sums
Suppose \(f(x)\) is positive and increasing.
On each subinterval, the left-endpoint value is the smallest function value on that interval.
Therefore, the left rectangle lies below the curve.
In this situation,
\[ \boxed{ L_n \text{ tends to underestimate the geometric area} }. \]Increasing Functions and Right Sums
For a positive increasing function, the right-endpoint value is the largest function value on each subinterval.
The right rectangle lies above most of the curve.
Therefore,
\[ \boxed{ R_n \text{ tends to overestimate the geometric area} }. \]Decreasing Functions Reverse the Pattern
If a positive function is decreasing, then:
- left-endpoint rectangles tend to overestimate;
- right-endpoint rectangles tend to underestimate.
The direction of change matters because it determines whether the selected endpoint is locally high or low.
Midpoint Estimates and Error Balance
Midpoint rectangles sample the interior of each subinterval.
Because they do not consistently use only the higher or lower endpoint, midpoint estimates often reduce systematic overestimation or underestimation.
They are still approximations, but they can be more accurate than equally coarse left or right sums.
More Rectangles Usually Improve the Approximation
Increasing \(n\) decreases the rectangle width:
\[ \Delta x=\frac{b-a}{n}. \]Narrower rectangles follow the curve more closely.
Thus, for a sufficiently well-behaved function,
\[ \boxed{ n\uparrow \Rightarrow \Delta x\downarrow \Rightarrow \text{typically better approximation} }. \]Example: Two Rectangles versus Four
For
\[ f(x)=x^2 \]on
\[ [0,2], \]a two-rectangle left sum uses
\[ \Delta x=1. \]The left endpoints are \(0\) and \(1\), so
\[ L_2 = 1[0^2+1^2] = \boxed{1}. \]With four rectangles, we found
\[ L_4=1.75. \]The refined estimate moves closer to the true area.
Numerical Area from a Table
Rectangle methods do not require a formula.
Suppose equally spaced data are given:
| \(x\) | \(f(x)\) |
|---|---|
| \(0\) | \(2\) |
| \(1\) | \(3\) |
| \(2\) | \(5\) |
| \(3\) | \(6\) |
| \(4\) | \(8\) |
With width \(1\), a left sum over \([0,4]\) is
\[ L_4 = 1(2+3+5+6) = \boxed{16}. \]Right Sum from the Same Table
Using the right endpoints,
\[ R_4 = 1(3+5+6+8) = \boxed{22}. \]Because the data are increasing, the left estimate is lower and the right estimate is higher.
Midpoint Data Requirements
A midpoint estimate requires function values at subinterval midpoints.
If the table contains only endpoint values, midpoint data may be unavailable.
In that case, we should not invent midpoint heights unless the problem explicitly allows interpolation or provides another model.
Unequal Subinterval Widths
Rectangle approximation also works when subinterval widths are unequal.
If the widths are
\[ \Delta x_1,\Delta x_2,\ldots,\Delta x_n, \]then the estimate becomes
\[ \boxed{ \sum_{i=1}^{n} f(x_i^*)\Delta x_i }, \]where \(x_i^*\) is the chosen sample point in the \(i\)th subinterval.
Signed Rectangle Contributions
If the function value is negative, the corresponding rectangle contributes negative signed area.
For example, if
\[ f(x_i^*)=-3 \]and
\[ \Delta x=2, \]then the contribution is
\[ (-3)(2) = \boxed{-6}. \]Example: Signed Numerical Accumulation
Suppose a rate is sampled at left endpoints with values
\[ 4,\ 2,\ -1,\ -3 \]and each interval has width
\[ 2. \]The left approximation is
\[ 2(4+2-1-3). \] \[ =2(2) = \boxed{4}. \]Positive and negative contributions combine to produce net accumulated change.
Total Geometric Area from Rectangles
If the goal is total geometric area rather than signed accumulation, use the magnitudes of the rectangle contributions.
In the previous example,
\[ 2(|4|+|2|+|-1|+|-3|) \] \[ = 2(10) = \boxed{20}. \]Signed accumulation and total area answer different questions.
Rectangle Sums as Rate Accumulation
Suppose \(r(t)\) is a rate.
On a short interval of width \(\Delta t\), approximate the rate by a representative value \(r(t_i^*)\).
The accumulated contribution is approximately
\[ r(t_i^*)\Delta t. \]Summing gives
\[ \boxed{ \sum r(t_i^*)\Delta t }. \]This is the rate-based interpretation of a rectangle sum.
Example: Velocity Data
Suppose velocity is recorded every second:
| \(t\) | \(v(t)\) |
|---|---|
| \(0\) | \(2\) |
| \(1\) | \(3\) |
| \(2\) | \(5\) |
| \(3\) | \(4\) |
| \(4\) | \(1\) |
A left-endpoint estimate of displacement from \(t=0\) to \(t=4\) is
\[ 1(2+3+5+4) = \boxed{14}. \]If velocity is measured in meters per second, the estimated displacement is \(14\) meters.
Right-Endpoint Estimate from the Velocity Data
The right-endpoint estimate is
\[ 1(3+5+4+1) = \boxed{13}. \]The two estimates differ because the velocity changes during the interval.
Average of Left and Right Estimates
A simple way to combine left and right information is to average the two estimates:
\[ \frac{L_n+R_n}{2}. \]For the velocity example,
\[ \frac{14+13}{2} = \boxed{13.5}. \]This averaging idea is closely related to the trapezoidal rule, which will be developed more systematically in numerical methods.
Connection to Trapezoidal Approximation
On one subinterval, averaging the left and right heights gives
\[ \frac{f(x_i)+f(x_{i+1})}{2}. \]Multiplying by the width gives
\[ \frac12 \left[ f(x_i)+f(x_{i+1}) \right]\Delta x. \]This is exactly the area formula for a trapezoid with vertical parallel sides.
Riemann-Sum Notation as a Preview
A general rectangle sum can be written
\[ \boxed{ \sum_{i=1}^{n} f(x_i^*)\Delta x }. \]Such sums are called Riemann sums.
The notation \(x_i^*\) means that a sample point has been chosen in the \(i\)th subinterval.
Left, right, and midpoint sums are special cases of this general form.
From Riemann Sums to Definite Integrals
If the partition becomes finer and the rectangle widths approach zero, the Riemann sums may approach a stable value.
That limiting value is the definite integral:
\[ \boxed{ \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\Delta x } \]when the limit exists.
At this stage, the formula should be understood as the mathematical ideal of increasingly fine rectangular approximation.
Numerical Approximation versus Exact Integration
A rectangle sum with finitely many rectangles is generally an approximation.
The definite integral is the exact limiting accumulation when it exists.
Numerical methods remain important even after antiderivatives are introduced because many real data sets and many functions do not have simple elementary antiderivative formulas.
Error Depends on More Than the Number of Rectangles
Using more rectangles often improves an estimate, but accuracy also depends on:
- how rapidly the function changes;
- the chosen sample points;
- the interval width;
- the smoothness of the function;
- whether the graph bends strongly within subintervals.
Numerical approximation is a balance between simplicity and accuracy.
Example: Refinement Pattern
Suppose a sequence of midpoint estimates is
\[ M_2=2.50, \] \[ M_4=2.625, \] \[ M_8=2.65625. \]The values appear to stabilize near
\[ 2.67. \]This kind of stabilization is numerical evidence that the approximations are converging toward a limiting area.
Overestimate and Underestimate Are Contextual
The usual left-under/right-over rule applies to positive increasing functions.
If the function changes direction or crosses the axis, the relationship may vary from one subinterval to another.
Therefore, do not apply a memorized overestimate rule without checking the graph's actual behavior.
Rectangle Methods Work with Measured Data
One major advantage of numerical accumulation is that it can be used when no symbolic formula is known.
If measurements of a rate are available at known times, each measurement can serve as a rectangle height.
This makes rectangular approximation useful in experiments, finance, engineering, environmental studies, and other data-driven settings.
Common Error: Using the Wrong Number of Widths
If \(n\) rectangles cover
\[ [a,b], \]then
\[ \Delta x=\frac{b-a}{n}. \]The number of partition points is usually
\[ n+1. \]Confusing rectangles with partition points produces the wrong width.
Common Error: Using Both Endpoints in a Left Sum
For \(n\) subintervals, a left sum uses \(n\) left endpoints:
\[ x_0,x_1,\ldots,x_{n-1}. \]It does not use the final endpoint \(x_n\).
A right sum instead uses
\[ x_1,x_2,\ldots,x_n. \]Common Error: Forgetting the Width Factor
The expression
\[ f(x_0)+f(x_1)+\cdots \]adds heights, not areas.
Each height must be multiplied by its interval width.
For equal widths,
\[ \boxed{ \text{sum} = \Delta x \sum \text{heights} }. \]Common Error: Treating a Negative Height as Positive
For signed accumulation, a negative function value gives a negative rectangle contribution.
Replacing all heights by positive values changes the problem from net accumulation to total geometric area.
Common Error: Calling a Finite Rectangle Sum Exact
Unless the function is especially simple on each subinterval or the problem states otherwise, a finite rectangle sum is an approximation.
Use notation such as
\[ A\approx L_n \]or
\[ A\approx M_n \]when appropriate.
Common Error: Assuming Midpoint Data Exist
A midpoint sum requires midpoint function values.
If only endpoint data are given, those midpoint values may not be known.
Do not silently create data that the problem has not supplied.
Common Error: Ignoring Units
If the height represents
\[ \frac{\text{liters}}{\text{minute}} \]and the width represents minutes, then each rectangle area has units
\[ \text{liters}. \]Numerical accumulation should always be interpreted through its units.
A Reliable Procedure for Equal-Width Rectangle Sums
- Identify the interval \([a,b]\).
- Identify the number of rectangles \(n\).
- Compute \(\Delta x=(b-a)/n\).
- List the partition points.
- Choose left, right, or midpoint sample points as required.
- Evaluate the function at those sample points.
- Add the selected heights.
- Multiply the sum by \(\Delta x\).
- Preserve signs for signed accumulation.
- Attach the correct units and approximation notation.
A Reliable Procedure for Table-Based Accumulation
- Inspect the input spacing.
- Determine whether the widths are equal or unequal.
- Select the required endpoint or sample values.
- Multiply each sampled rate or height by its corresponding interval width.
- Add all signed contributions.
- If total geometric area is required, use magnitudes instead.
- Interpret the result in the context of the data.
Example: Full Left-Sum Workflow
Estimate the area under
\[ f(x)=x+1 \]on
\[ [0,4] \]using \(4\) left-endpoint rectangles.
Width:
\[ \Delta x = \frac{4-0}{4} = 1. \]Left endpoints:
\[ 0,\ 1,\ 2,\ 3. \]Heights:
\[ 1,\ 2,\ 3,\ 4. \]Therefore,
\[ L_4 = 1(1+2+3+4) = \boxed{10}. \]Compare with the Right Sum
The right endpoints are
\[ 1,\ 2,\ 3,\ 4. \]The heights are
\[ 2,\ 3,\ 4,\ 5. \]Therefore,
\[ R_4 = 1(2+3+4+5) = \boxed{14}. \]Since the function is increasing, the true area lies between these two endpoint estimates.
Midpoint Estimate for the Same Function
The midpoints are
\[ 0.5,\ 1.5,\ 2.5,\ 3.5. \]Their function values are
\[ 1.5,\ 2.5,\ 3.5,\ 4.5. \]Therefore,
\[ M_4 = 1(1.5+2.5+3.5+4.5) = \boxed{12}. \]Because \(f(x)=x+1\) is linear, this midpoint estimate equals the exact trapezoidal area.
Practice 1: Find \(\Delta x\)
Divide
\[ [2,10] \]into \(4\) equal subintervals.
Find \(\Delta x\).
Practice 2: Left Sum
For
\[ f(x)=x+2 \]on
\[ [0,4], \]find \(L_4\).
Practice 3: Right Sum
For the same function and interval, find \(R_4\).
Practice 4: Midpoint Sum
For
\[ f(x)=x+2 \]on
\[ [0,4], \]find \(M_4\).
Practice 5: Increasing Function Estimate
A positive function is increasing on an interval.
Which tends to underestimate the area: the left-endpoint sum or the right-endpoint sum?
Practice 6: Signed Rectangle Sum
Four left-endpoint heights are
\[ 3,\ 1,\ -2,\ -4 \]and each subinterval has width \(2\).
Find the signed accumulation.
Practice 7: Total Geometric Area Estimate
Using the same heights and widths as Practice 6, find the total geometric rectangle area.
Practice 8: Table-Based Left Sum
| \(x\) | \(f(x)\) |
|---|---|
| \(0\) | \(4\) |
| \(2\) | \(5\) |
| \(4\) | \(7\) |
| \(6\) | \(8\) |
Use a left-endpoint sum to estimate accumulation from \(x=0\) to \(x=6\).
Practice 9: Refinement
On a fixed interval, what happens to
\[ \Delta x \]when the number of equal rectangles \(n\) increases?
Practice 10: Identify the Error
A student divides \([0,12]\) into \(4\) rectangles and writes
\[ \Delta x=\frac{12}{5}. \]Explain the error and give the correct width.
Practice Answers
Practice 1
\[ \Delta x = \frac{10-2}{4} = \frac{8}{4} = \boxed{2}. \]Practice 2
Here,
\[ \Delta x=1. \]Left endpoints:
\[ 0,\ 1,\ 2,\ 3. \]Heights:
\[ 2,\ 3,\ 4,\ 5. \]Therefore,
\[ L_4 = 1(2+3+4+5) = \boxed{14}. \]Practice 3
Right endpoints:
\[ 1,\ 2,\ 3,\ 4. \]Heights:
\[ 3,\ 4,\ 5,\ 6. \] \[ R_4 = 1(3+4+5+6) = \boxed{18}. \]Practice 4
Midpoints:
\[ 0.5,\ 1.5,\ 2.5,\ 3.5. \]Heights:
\[ 2.5,\ 3.5,\ 4.5,\ 5.5. \] \[ M_4 = 1(2.5+3.5+4.5+5.5) = \boxed{16}. \]Practice 5
For a positive increasing function, the
\[ \boxed{\text{left-endpoint sum}} \]tends to underestimate the area.
Practice 6
\[ 2(3+1-2-4) = 2(-2) = \boxed{-4}. \]Practice 7
\[ 2(|3|+|1|+|-2|+|-4|) \] \[ = 2(10) = \boxed{20}. \]Practice 8
Each width is
\[ 2. \]A left sum uses the heights
\[ 4,\ 5,\ 7. \]Therefore,
\[ L_3 = 2(4+5+7) = 2(16) = \boxed{32}. \]Practice 9
Since
\[ \Delta x=\frac{b-a}{n}, \]increasing \(n\) makes
\[ \boxed{\Delta x\text{ smaller}}. \]Practice 10
Four rectangles mean \(n=4\), not \(5\).
\[ \Delta x = \frac{12-0}{4} = \boxed{3}. \]What to Remember
Equal-width rectangular approximation begins with
\[ \boxed{ \Delta x=\frac{b-a}{n} }. \]Then
\[ \boxed{ \text{rectangle estimate} = \Delta x \sum \text{sampled heights} }. \]- left sums use left endpoints;
- right sums use right endpoints;
- midpoint sums use subinterval midpoints;
- positive increasing functions tend to have left underestimates and right overestimates;
- decreasing functions reverse that endpoint pattern;
- negative heights contribute negative signed area;
- absolute values are used when total geometric area rather than net accumulation is required;
- more and narrower rectangles generally improve approximation;
- rectangle methods work with formulas or measured data;
- Riemann sums provide the limiting bridge from finite rectangles to definite integrals.
Where We Go Next
Rectangular approximation shows how accumulated change can be built from many small contributions. The definite integral represents the limiting value of this process. For many simple functions, however, we can compute the same accumulation much more efficiently by reversing differentiation.
The next lesson, Simple Antiderivatives and the Rate–Accumulation Connection, introduces antiderivatives and develops the fundamental relationship between local rate and total accumulated change.