Using Rectangles to Bound an Integral
A definite integral of a nonnegative function represents accumulated area. Often, a graph or formula gives enough information to estimate that area even when an exact calculation is unavailable. Rectangles provide a direct geometric method: rectangles below the graph give a lower bound, while rectangles above it give an upper bound.
The essential task is to choose each rectangle’s height carefully. Its width comes from the chosen interval, but its height must be no greater than the function throughout that interval for a lower estimate, or no less than the function throughout it for an upper estimate.
Writing \(\Delta x_i=x_i-x_{i-1}\), the total area of the lower rectangles is \(\sum_{i=1}^{n}h_i\Delta x_i\); the total area of the upper rectangles is \(\sum_{i=1}^{n}k_i\Delta x_i\). When all the required pointwise bounds hold, these sums bound the integral:
Why the Rectangle Bounds Work
The justification follows from integral comparison, interval additivity, and the integral of a constant function. On a given subinterval, the constant function with value \(h_i\) lies below \(f\), so its integral cannot exceed the integral of \(f\) there. The constant function with value \(k_i\) lies above \(f\), so its integral cannot be smaller. The integral of a constant is its height times the interval width.
This argument also explains the geometric interpretation. A rectangle below the graph contributes no more area than the graph contributes over the same interval. A rectangle above the graph contributes at least as much. Combining the rectangles combines these comparisons across the whole interval.
Worked Examples
Worked Example: Endpoint Rectangles for an Increasing Line
Estimate \(\int_0^2 (x+1)\,dx\) using two equal subintervals. The function \(f(x)=x+1\) is increasing, so on each subinterval its left-endpoint value is a lower height and its right-endpoint value is an upper height.
The partition points are \(x_0=0\), \(x_1=1\), and \(x_2=2\). Both widths are \(1\). On \([0,1]\), the endpoint values are \(f(0)=1\) and \(f(1)=2\), giving lower and upper rectangle areas \(1\cdot1=1\) and \(2\cdot1=2\). On \([1,2]\), the values are \(f(1)=2\) and \(f(2)=3\), giving areas \(2\cdot1=2\) and \(3\cdot1=3\).
Adding the areas gives the bounds:
Check using an antiderivative: \(\int_0^2(x+1)\,dx=[x^2/2+x]_0^2=(2+2)-0=4\). Thus \(3\le4\le5\), as required.
Worked Example: More Rectangles for a Curved Graph
Estimate \(\int_0^2 x^2\,dx\) using four subintervals of width \(1/2\). Since \(x^2\) is increasing on \([0,2]\), the left endpoint gives a lower height and the right endpoint an upper height on each subinterval.
The partition points are \(0,\frac12,1,\frac32,2\). The lower heights are \(0^2=0\), \((\frac12)^2=\frac14\), \(1^2=1\), and \((\frac32)^2=\frac94\). The lower total area is therefore \(\frac12(0+\frac14+1+\frac94)=\frac12\cdot\frac72=\frac74\).
The upper heights are \((\frac12)^2=\frac14\), \(1^2=1\), \((\frac32)^2=\frac94\), and \(2^2=4\). The upper total area is \(\frac12(\frac14+1+\frac94+4)=\frac12\cdot\frac{15}{2}=\frac{15}{4}\). Thus:
Check using \(\int x^2\,dx=x^3/3\): \(\int_0^2x^2\,dx=8/3\). In common denominators, \(7/4=21/12\), \(8/3=32/12\), and \(15/4=45/12\), so \(21/12\le32/12\le45/12\). The integral lies within the rectangle bounds.
Worked Example: Bounding a Graph That Is Not Monotone
Consider \(f(x)=1+(x-1)^2\) on \([0,2]\). Its graph decreases to \(f(1)=1\) and then increases. The endpoint values are \(f(0)=2\) and \(f(2)=2\), but the function is not at least \(2\) throughout the interval. Instead, \(0\le(x-1)^2\le1\) for \(0\le x\le2\), so \(1\le f(x)\le2\) everywhere on the interval.
Use one lower rectangle of height \(1\) and one upper rectangle of height \(2\), each with width \(2\). Their areas are \(1\cdot2=2\) and \(2\cdot2=4\), respectively. Consequently:
Check the bounds exactly. An antiderivative is \(x+(x-1)^3/3\), so the integral is \((2+1/3)-(0-1/3)=8/3\). Since \(2=6/3\) and \(4=12/3\), we have \(6/3\le8/3\le12/3\).
Choosing Heights from Graph Information
For a monotone function, endpoint values often give convenient rectangle heights. If \(f\) is increasing on a subinterval, its left-endpoint value is no greater than any value on that interval, and its right-endpoint value is no less. For a decreasing function, the roles reverse: the right-endpoint value is a lower height, and the left-endpoint value is an upper height.
Monotonicity is one way to justify a height, but it is not the only way. A formula may show that the function stays between two constants, as in the preceding example. Graph information may also establish a lower or upper level across a whole interval. In every case, the justification must apply throughout the subinterval—not just at a single point.
Once valid heights have been chosen, keep the calculation in two stages: first find each rectangle’s area, then add the areas. For unequal subintervals, use each interval’s own width. For example, the contribution from \([x_{i-1},x_i]\) is \(h_i(x_i-x_{i-1})\), not a common width multiplied by every height unless the intervals really are equal in width.
A Common Pitfall: Endpoint Values Are Not Always Bounds
A tempting but unreliable shortcut is to use the values at the two ends of an interval as if they automatically describe the function everywhere between them. The nonmonotone example shows why this can fail. Both endpoint values are \(2\), but the middle value is \(f(1)=1\). Therefore, a rectangle of height \(2\) is not below the graph on \([0,2]\).
A second practical check is to confirm the dimensions and the direction of the inequality. Each area is height times width, so it is nonnegative here. A lower rectangle must yield a lower bound, and an upper rectangle must yield an upper bound. If the calculated areas do not have that order, revisit the height choices and arithmetic.
Why Rectangular Estimates Matter
Rectangle estimates translate geometric knowledge into precise statements about integrals. They are useful when an exact value is difficult to find or when only a graph, a table, or simple bounds on a function are available. A coarser partition may give a wide interval of possible values; using more subintervals can provide more opportunities to fit rectangles closely to the graph. The validity of the bounds, however, always depends first on choosing heights that genuinely bound the function.
The method can be summarized in three checks: identify the width of each interval, justify a lower or upper height over that whole interval, and add the resulting areas. When an exact integral is available, it provides a useful final check that the estimate is consistent.
Check Your Understanding
Use the rectangle method and the conditions stated in each question.
- For an increasing function on \([a,b]\), which endpoint value gives a lower rectangle height, and which gives an upper height?
- A rectangle has height \(3\) over an interval of width \(2/5\). What is its area?
- Explain why endpoint values alone do not necessarily establish a lower or upper bound for a function on an interval.
- Suppose \(2\le f(x)\le5\) for every \(x\in[1,4]\). Give one lower and one upper rectangular bound for \(\int_1^4 f(x)\,dx\).
- For a decreasing function on \([0,3]\), state which endpoint value gives the upper rectangle height.
