From rectangles to accumulated change
A rate that varies over time does not have one constant value to multiply by the whole time interval. Instead, divide the interval into smaller pieces, estimate the rate on each piece, and add the resulting changes. A rate that is positive contributes positive accumulation; a negative rate contributes negative accumulation.
Geometrically, each estimate is the signed area of a rectangle: its width is a subinterval length, and its height is a sampled value of the function. Adding the rectangles gives an approximation to the total accumulation. Making the rectangles narrower can improve the approximation, and the limiting value gives the idea of the definite integral.
Equal-width rectangular approximations
Let \(f\) be a function defined on a closed interval \([a,b]\), where \(a<b\). Divide the interval into \(n\) subintervals of equal width. The width of each is
In subinterval number \(i\), choose a point \(x_i^*\) and use \(f(x_i^*)\) as the rectangle's height. The total signed area of these \(n\) rectangles is
This number is a rectangular approximation to the function's accumulation on \([a,b]\). The points can be chosen by a rule, such as using the left endpoint, right endpoint, or midpoint of each subinterval. The chosen rule affects a finite approximation; it need not affect the limiting value.
For example, if a function is a rate measured in units per second and \(x\) is measured in seconds, then \(f(x_i^*)\Delta x\) has the units of the accumulated quantity. Units provide a useful check: adding the heights alone would not give accumulated change.
The definite integral as a limit
As \(n\) increases, the common width \(\Delta x\) decreases. If the resulting rectangular approximations approach one value, independently of the sample choices, that value is the function's total signed accumulation on the interval. This limiting value is called the definite integral of \(f\) from \(a\) to \(b\), and is written
The symbols \(a\) and \(b\) are the limits of integration. The \(dx\) indicates that the accumulation is with respect to changes in \(x\). The integral is a number, not a function of \(x\). Its sign reflects the net signed accumulation: portions where \(f\) is below the horizontal axis contribute negatively.
This definition expresses the central idea without requiring that a finite collection of rectangles match the graph exactly. An approximation is a finite sum; the integral is the limit those sums approach. For the elementary functions used below, we can calculate the sums and their limits directly.
Worked Example: A constant rate
Worked Example: A constant rate
Find the rectangular approximation for \(f(x)=3\) on \([-1,2]\), using \(n\) equal-width subintervals, and identify its limiting value.
The interval has length \(2-(-1)=3\), so each subinterval has width \(\Delta x=3/n\). Since the function has value \(3\) everywhere, every rectangle has height \(3\). Each contributes
There are \(n\) such contributions, so their total is
This value is the same for every \(n\), so the limit is \(9\). The graph is a horizontal line at height \(3\), and the rectangle covering the interval has width \(3\) and height \(3\), giving signed area \(3\cdot3=9\). This geometric check agrees with the rectangular sum.
Worked Example: Left rectangles for \(f(x)=x\)
Worked Example: Left rectangles for \(f(x)=x\)
Approximate the accumulation of \(f(x)=x\) on \([0,1]\) using left endpoints, then find the limit as the number of subintervals increases.
Each subinterval has width \(1/n\). The left endpoints are \(0,1/n,2/n,\ldots,(n-1)/n\). Thus the left-endpoint sum is
The sum of the integers from \(0\) through \(n-1\) is \(n(n-1)/2\). Therefore
As \(n\) increases without bound, \(1/(2n)\) approaches \(0\), so \(L_n\) approaches \(1/2\).
For a check, take \(n=4\). The width is \(1/4\), and the left-endpoint heights are \(0,1/4,1/2,3/4\). Their total signed area is
The formula gives \(L_4=(4-1)/(2\cdot4)=3/8\), as required. This finite approximation is below the limiting value \(1/2\), which is consistent with the increasing graph: on each subinterval the left endpoint gives a height no greater than the function's values to its right.
Why the sample choice can disappear in the limit
The previous example also illustrates why the left-endpoint rule need not determine the final answer. Since \(f(x)=x\) is increasing on \([0,1]\), the left rectangle on each subinterval is no taller than the function there, and the right rectangle is no shorter. Consequently, any choice of sample point within each subinterval produces a total between the left- and right-endpoint totals.
For this function, the right-endpoint sum is
Every sample-point sum lies between \(L_n\) and \(R_n\), and the bounds approach the same number:
Both bounds approach \(1/2\), so the sample-point sums are forced to approach \(1/2\) as well. This is a complete reason, for this example, that changing the sample rule does not change the limiting value.
Worked Example: Midpoint rectangles for \(f(x)=x^2\)
Worked Example: Midpoint rectangles for \(f(x)=x^2\)
Find the limit of the midpoint-rectangle sums for \(f(x)=x^2\) on \([0,2]\).
The width of each of \(n\) equal subintervals is \(2/n\). Number the subintervals from \(i=0\) to \(i=n-1\). Their midpoints are \( (2i+1)/n\): for example, the first is \(1/n\), and the last is \((2n-1)/n\). The midpoint sum is therefore
To evaluate the bracketed sum, expand the odd squares. The odd numbers are \(2i+1\) for \(i=0,\ldots,n-1\), so their squares add to
Here the standard finite-sum formulas are \(0+1+\cdots+(n-1)=n(n-1)/2\) and \(0^2+1^2+\cdots+(n-1)^2=n(n-1)(2n-1)/6\). Substituting the odd-square sum gives
Since \(2/(3n^2)\) approaches zero, the midpoint sums approach \(8/3\). Check the finite calculation at \(n=2\): the midpoints are \(1/2\) and \(3/2\), the width is \(1\), and the sum is
The formula gives \(M_2=8/3-2/(3\cdot4)=8/3-1/6=5/2\), confirming the calculation. As the rectangles become narrower, the correction term \(2/(3n^2)\) vanishes, leaving the limiting accumulation \(8/3\).
Signed area is essential
A rectangle below the horizontal axis has a negative height when its height is taken from the function value. Its contribution to the sum is therefore negative. This is necessary when the quantity being accumulated is a signed change, such as displacement from a velocity function or net flow from a rate that can reverse direction.
For a concrete example, consider \(f(x)=x-1\) on \([0,2]\). The graph lies below the axis from \(0\) to \(1\), and above it from \(1\) to \(2\). The two triangular regions have areas \(1/2\) each, but their signed contributions are \(-1/2\) and \(+1/2\), so the net accumulation is \(0\).
The distinction also appears in finite sums. With \(n\) equal subintervals and left endpoints, the width is \(2/n\), and the sample points are \(2i/n\) for \(i=0,\ldots,n-1\). The signed sum is
This approaches \(0\), matching the signed-area calculation. Replacing every negative height by its absolute value would prevent the cancellation and would not compute the definite integral.
What the approximation viewpoint gives us
Rectangular sums connect a local rate to a total change. Each rectangle estimates the change over a short piece of the interval; adding the pieces estimates the whole change. The definite integral captures the value approached as those pieces become arbitrarily narrow.
Three checks help keep this interpretation clear:
- Width: The subinterval widths must add to \(b-a\).
- Units: Height times width must have the units of the accumulated quantity.
- Sign: Contributions below the axis remain negative when computing net accumulation.
A finite sum is useful even when the limiting value is not yet known, because it makes the approximation and its assumptions visible. When a limit can be found, it replaces the sequence of estimates with one exact accumulated value.
Check Your Understanding
Answer each question using signed rectangle sums and limiting values.
- For \(f(x)=2\) on \([1,4]\), what is the contribution of one rectangle when there are \(n\) equal subintervals? What is the total?
- For \(f(x)=x\) on \([0,1]\), why does every sample-point sum lie between the left- and right-endpoint sums?
- What value do the left-endpoint sums for \(f(x)=x\) on \([0,1]\) approach?
- For \(f(x)=x-1\) on \([0,2]\), why is the signed accumulation different from the total unsigned area?
- What does the definite integral represent in terms of rectangular approximations?
