6.3 The Definite Integral: Numerical and Graphical Approaches

(This topic is also in Section 6.3 in Applied Calculus and Section 13.3 in Finite Mathematics and Applied Calculus)


Numerical Integration Utility     Excel Riemann Sum Grapher
Graphing Calculator Programs     Online Text: Numerical Integration

In the textbook we consider an example (Example 1) based in which you are given the marginal cost of making a cell phone call and need to compute the total cost of a two hour call. Instead of going over the same example here, we look at a slightly different scenario here:

The Main Example: A pump is delivering water into a tank at a rate of

where t is time in minutes since the pump is turned on. What is the total amount of water pumped into the tank in the first two minutes?

First Rough Calculation: Since r(t) is the number of liters per minute at which water is entering the tank, when the pump is turned on, this rate is

so the total amount of water pumped into the tank in the first two minutes should be about:

Here is a graph showing that calculation we just did:

Total amount pumped in = 5 liters/minute × 2 Minutes

Notice two things:

Referring t the second point above, the reason that we have underestimated the total amount pumped in is that we used the initial rate r(0) throughout the entire 2-minute period.

Let us now redo the calculation using a minute-by-minute calculation, using the rate at the start of each minute for the calculation for that minute:

This gives a more accurate estimate of the total amount pumped in as

The new calculation is shown in the following graph:

Minute 1: 5 liters/minute × 1 Minute = 5 liters
Minute 2: 8 liters/minute × 1 Minute = 8liters

Notice that the new estimate is given by the sum of the areas of the two red rectangles.

But wait! This answer is still not exact: For instance, during Minute 1, the rate does not remain constant at 5 liters/minute, but is increasing there as well, and during the second minute, the rate is not constant at 8 liters/minute either. So we will next compute a more accurate third answer by using half-minute by half-minute calculations. But before we go on, let is write down what we have computed using Δt to denote the length of time we use in each step of the computation:

Each of these calculations is called a left Riemann sum of the function r(t); the first calculation is a Riemann sum with 1 subdivision, the second a Riemann sum with 2 subdivisions, and the third Riemann sum with 4 subdivisions.

Here is the graph for the third calculation:

Riemann Sum with 4 subdivisions:      
r(0)Δt + r(0.5)Δt + r(1)Δt + r(1.5)Δt = 15.25 liters

To compute the Riemann sum with 5 subdivisions, use rectangles of width Now we calculate the height of each rectangle:

To obtain the Riemann sum, we add the areas of all five rectangles. The area of each rectangle is its height times its width Δt (0.4). Instead of multiplying each height by 0.4 and then adding, we can equivalently add the heights and multiply the sum by 0.4:

If you want to see Riemann sums graphically for different numbers of subdivisions, go to the Excel Riemann Sum Grapher, make sure that macros are enabled in Excel (If your Excel security is set to "high," macros will be automatically disabled and the utility will not work, so you will have to reset the security to "medium" through the "Tools" menu in Excel and then relaunch Excel.) Enter 3*x^2+5 in the Function box, set "a" and "b" to 0 and 2 respectively, and choose the number of subdivisions you want to use.

Riemann Sum

If f is a function defined on the interval [a, b] and n is a number (of subdivisions), we define the associated left Riemann sum with n equal subdivisions as follows:

  • Subdivide the interval [a, b] into n equal intervals of width Δx = (b - a)/n.
  • Start with x0 = a and successively add Δx to get additional points
    x1 = x0 + Δx;
    x2 = x1 + Δx;
    xn-1 = xn-2 + Δx.
  • Calculate [f(x0) + f(x1) + f(x2) + ... + f(xn-1)]Δx

    This is the desired Riemann sum.


  1. We just calculated several Riemann sums for the function f(x) = 3x2 + 5 (although we used different letters for the name of the function and the independent variable.) Check to see how the calculations we did match the formula above.
  2. Let f(x) = x2, and let [a, b] = [-1, 1]. The Riemann sum for n = 4 subdivisions is calculated as follows:
    • Δx =      
    • x0 =
      x1 =
      x2 =
      x3 =    
    • Riemann Sum =      

Let us go back to thinking about the Main Example of water being pumped into a tank.

Q None of the Riemann sums we calculated for the water being pumped into the tank gave the right answer; in fact all these calculations ignore the fact that the rate keeps increasing over each interval, so Riemann sums are a waste of time, right?
A The first claim is right, but the second claim is not: Each calculation gives us a more accurate estimate of the amount of water pumped in, and in fact we have discovered an astounding fact:

The exact amount of water pumped in is: (There are two correct answers. Try to click on the two correct ones in succession.)

We have a name for this limit of the Riemann sums or, equivalently, the area under a curve above an interval:

Definite Integral

If f is a function defined on the interval [a, b] then the definite integral of f over [a, b] is defined to be the limit of the Riemann sums as the number of subdivisions n → ∞, assuming the limit exists. We use the following notation for the definite integral:

    Definite Integral of f over [a, b]=Limit of Riemann Sums as n → &infin

    f(x) dxDefinite integral, from a to b, of f(x) with respect to x

Geometrically, the definite integral measures area. If the graph of the function is above the x-axis, the area is counted as positive (like the examples we have been looking at). If the graph of the function is below the x-axis, the area is counted as negative. The following picture illustrates the general case:


f(x) dx=Area A - Area B

Interpretation of the Definite Integral
If f is the rate of change of a quantity F (that is, f = F'), then b

f(x) dxis the (exact) total change of F from x = a to x = b.


We can sometimes calculate the definite integrqal geometrically by calculating the area under the curve directly as the following examples show:


    2 dx=

    (1+x) dx=

    f(x) dx=

Q What about areas that are not easy to calculate geomterically?
A The Funcdamental Theorem of Calculus (next section) tells us how to calculate these using antiderivatives. Otherwise, we can alsways use the numerical integration utility to calculate Riemann sums for larger and larger numbers of subdivisions to estimate the integral, as we will in the next example:

A projectile has velocity v(t) = 300 + 5t - 10t2 m/s after t seconds after being fired. Use Riemann sums for several values of n to estimate the total distance covered by the projectile from time t = 2 seconds to time t = 4 seconds.

Q How on earth am I supposed to find the distance covered knowing only the (varying) speed?
A Think for a moment about the pump example with which we started: We discovered that:

The total amount of water pumped in over the interval [0, 2] equals the definite integral of the rate from 0 to 2.

More generally, we have (see the "Interpretation" in the above box):

The total change in F over the interval [a, b] equals the definite integral of the rate of change of F from a to b.

In the situation we have here, the velocity v(t) is rate of change of position. So, the total change in position over the interval [2, 4] is

Since v(t) = 300 + 5t - 10t2, we have to compute4

(300 + 5t - 10t2) dt,which can do numerically using the numerical integration utility.

To use the utility, click on the above link, enter the function 300 + 5x - 10x^2 in the "f(x) = " box, enter 2 and 4 in the "Left End-Point" and "Right End-Point" boxes respectively, and try various values of n in the "Number of subdivisions" box. Then press "Left Sum" to get the left Riemann sum.

We conclude that the total distance covered (to the nearest meter) is about meters      

You now have several options

Top of Page

Last Updated: August, 2006
Copyright © 2006 Stefan Waner