Properties of Integrals

In the last two posts we looked at activities that lead to the FTC. Today I would like to build on that activity to demonstrate some of the important properties of integrals. I am never in favor of giving students theorems or properties like these and saying, “Here they are, learn them!” Nor do I think everything has to be done up with a detailed “proof.” So here is an approach between those alternatives.

Look at the activity in the last post where we considered three functions: F1(t) = 3, F2(x) = 2t and F3(t) = 2t + 3. We built new functions that gave the area between these functions and the t-axis. We expressed these as regular looking functions and also as functions defined by definite integrals – integrals where the upper limit of integration was a variable, x. We also made tables of values for these functions. Return to these tables and consider the following properties of integrals.

      1. Integrals are additive. I hope you noticed that F3 = F1 + F2.  Look at the values in the table you made for the area functions, and you will see that \int_{a}^{x}{F1(t)\pm F2(t)dt}=\int_{a}^{x}{F1(t)dt}\pm \int_{a}^{x}{F2(t)dt}. This is important because when evaluating definite integrals this allows us to do them term by term.
      2. Constants may be factored out of the integrand and written in front: \int_{a}^{b}{k\cdot f\left( t \right)}dt=k\cdot \int_{a}^{b}{f\left( t \right)dt} . To see this make a quick table for the area between F(t) = t and compare it to the area functions for F2(t) = 2t.
      3. Previously, we restricted our domain to nonnegative numbers. Let’s change that. Make a table for the three area functions A1(x), A2(x) and A3(x) for x = -1, -2, -3, -4, and -5. Work from the functions you wrote, not from the geometry. What do you notice? There are negative and zero values!  Areas are never negative. How do you explain what’s happened?

        Write the corresponding definite integrals. Be careful, the upper limits are now less than the lower limits. Think of the Riemann sums that are behind the limits. Since x moves left, the values in the Riemann sum will be negative producing negative values for A1(x). In A2(x) both and the function values are negative, producing a positive result. Can you explain what’s going on with A3(x) which has positive, zero and negative values?

        The short answer is that if the upper and the lower limits of integration are switched then the resulting integrals are opposites of each other:\int_{a}^{b}{f\left( x \right)dx}=-\int_{b}^{a}{f\left( x \right)dx}

      4. Next use your graph and geometry to find that area between F3(t) and the t-axis between, say, x = 3 and x =  5, and then use the table value to show that \int_{0}^{3}{F3(t)dt}+\int_{3}^{5}{F3(t)dt}=\int_{0}^{5}{F3(t)dt} . Now use those numbers and the property in paragraph 3 to show that \int_{3}^{5}{F3(t)dt}+\int_{5}^{0}{F3(t)dt}=\int_{3}^{0}{F3(t)dt}.
        In general, \int_{a}^{b}{f\left( x \right)dx}+\int_{b}^{c}{f\left( x \right)dx}=\int_{a}^{c}{f\left( x \right)dx} regardless of the order of a ,b and c. The only thing that matters is that (1) the lower limit in the first integral on the left is the lower limit on the right, (2) the upper limit on the last integral on the left is the upper limit on the right, and (3) on the left the upper limit on one integral is the lower limit on the next. You can even string more integrals together as long as you follow the pattern.
      1. A final property which can be seen by comparing the graphs and the areas between them and the t-axis for F2(x) and F3(x) is this: If f\left( x \right)\le g\left( x \right) on the interval [a, b], then \int_{a}^{b}{f\left( x \right)dx}\le \int_{a}^{b}{g\left( x \right)dx}. This is sometimes called the “Racetrack Principle.” Interpreting f and g as rates and their integrals as amounts (or distances), then in the same interval, the faster horse travels farther.

Happy Holiday. My next post will be Friday December 28, 2012.


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