Using well-known probability density functions to evaluate certain integrals in calculus.

``` ABSTRACT

A major source of anxiety for engineering and science majors who
must take calculus is the topic of integration. The techniques of
integration necessary for solving practical problems in the
sciences are lengthy and complex. In this paper, we are
introducing a method that minimizes the algebra as well as the
number of computations for certain types of integrals. It is
important to note that we are not suggesting that the conventional
methods for teaching techniques of integration be replaced with
the innovation we are proposing. Rather, the new approach we
suggest is to be presented only after the traditional techniques
of integration by parts and substitution have been taught. The new
approach is called the probability density function (p.d.f.)
method and uses the properties of probability density functions to
evaluate integrals for certain classes of functions. In this paper
we demonstrate that, for certain classes of functions, this method
greatly reduces the length and complexity of many integration
problems.

Key words: Probability Density Function (p.d.f.), Calculus,
Integration Techniques, Gamma Function, Beta Function, Cauchy
Function, Normal function
```

INTRODUCTION

Learning and practicing integration techniques is a standard requirement in the second semester calculus course. Unfortunately, some methods of integration can be difficult and time consuming for students, especially the method known as integration by parts. While introduced to calculus students as a means of obtaining a simpler, solvable integral from a more complex integral, integration by parts often involves multiple applications of the method in order to get the problem into a form simple enough to solve. The lengthy and tedious process involved in integration by parts often generates algebra errors that are hard to correct, leaving students frustrated with the method. Students may even find themselves asking if there is a better way to calculate some of the integrals requiring integration by parts. Fortunately, there is a better way. Some integrals can be evaluated using the properties of probability density functions for continuous random variables, thus simplifying and shortening the work.

A probability density function (p.d.f.) is a function defined on an interval (a, b) and having both of the following properties:

(i) f(x)[greater than or equal to]0 for every x, and

(ii) a.[integral].b f(x)dx = 1.

In (ii) it is important to note that a and/or b can be infinite, creating an improper integral. These properties of p.d.f.'s would likely be unfamiliar to the typical calculus student since p.d.f.'s are usually not taught in calculus courses. However, most statistics students are very familiar with the properties of p.d.f.'s.

One p.d.f. that is especially useful in calculus is the gamma function (3), which is given by f(x) = 1/[GAMMA]([alpha])[[beta].sup.[alpha]] [x.sup.[alpha]-1][e.sup.-x/[beta]]; 0<x<[infinity]; where[alpha]>0, [beta]>0; and [GAMMA]([alpha]) = ([alpha]-1)! and is denoted by X~[GAMMA]([alpha], [beta]). Since the gamma function (3) is a p.d.f., it integrates to one for all values of x, giving [infinity].[integral].0 1/[GAMMA]([alpha])[[beta].sup.[alpha]] [x.sup.[alpha]-1][e.sup.-x/[beta]] dx = 1; where[alpha]>0, [beta]>0; and [GAMMA]([alpha]) = ([alpha]-1)!.

Calculus books contain numerous integrals of this type, which can be evaluated using the properties of the gamma function. However, students are not taught to solve this type of problem by using the gamma function. Instead, students are taught to use integration by parts, which is a much more complicated and time consuming than applying the properties of p.d.f.'s. Following is an example of how the properties of the gamma function (3) can be applied to solve an integral faster than by using the method of integration by parts.

THE USE OF P.D.F.'S

To evaluate [infinity].[integral].0 [x.sup.2][e.sup.-x/2]dx using the properties of the gamma function (3), we have to introduce the correct constant, which is always 1/[GAMMA]([alpha]) * [[beta].sup.[alpha]]; this forces the integral to become a p.d.f. and integrate to one. First, rewrite the integral to match the gamma function, [infinity].[integral].0 [x.sup.3-1][e.sup.-x/2]dx. Hence, [alpha] = 3 and [beta] = 3, and the correct constant is 1/[GAMMA]([alpha])[[beta].sup.[alpha]] = 1/[GAMMA](3)[2.sup.3]. Therefore, [infinity].[integral].0 1/[GAMMA](3) * [2.sup.3][x.sup.3-1][e.sup.-x/2] dx = 1 since X ~ [GAMMA] ([alpha] = 3, [beta] = 3) Note that since we introduce the constant 1/[GAMMA](3)[2.sup.3], we have to multiply by [GAMMA] (3) * [2.sup.3] to cancel it out, i.e. [GAMMA] (3) * [2.sup.3] [infinity].[integral].0 1/[GAMMA](3)[2.sup.3][x.sup.3-1][e.sup.-x/2] dx. Now, evaluating the integral gives: [infinity].[integral].0 [x.sup.3-1][e.sup.-x/2] dx = [GAMMA](3) * [2.sup.3] [[infinity].[integral].0 1/[GAMMA](3)[2.sup.3][x.sup.3-1][e.sup.-x/2] dx] = [GAMMA](3) * [2.sup.3] * [1] = (3-1)!(8) = 16 Now to evaluate [infinity].[integral].0 [x.sup.2][e.sup.-x/2] dx using integration by parts, let u = [x.sup.2]; du = 2x dx; dv = [e.sup.-x/2] dx; and v = -2[e.sup.-x/2]. Now, [infinity].[integral].0 [x.sup.2][e.sup.-x/2] dx = lim.b[right arrow][infinity] b.[integral].0 [x.sup.2][e.sup.-x/2] dx = lim.b[right arrow][infinity](uv[|.sub.0.sup.b] - b.[integral].0 v du) = lim.b[right arrow][infinity](-2[x.sup.2][e.sup.-x/2][|.sub.0.sup.b] - b.[integral].0 -4x[e.sup.-x/2] dx).

At this point, the integral still cannot be evaluated and another application of integration by parts must be applied, this time to the new integral b.[integral].0 -4x[e.sup.-x/2] dx

Letting u = -4x; du = -4 dx; dv = [e.sup.-x/2] dx; v = -2[e.sup.-x/2] gives lim.b[right arrow][infinity](-2[x.sup.2][e.sup.-x/2][|.sub.0.sup.b] - b.[integral].0 -4x[e.sup.-x/2] dx) = lim.b[right arrow][infinity](-2[x.sup.2][e.sup.-x/2][|.sub.0.sup.b] - (8x[e.sup.-x/2][|.sub.0.sup.b] - b.[integral].0 8[e.sup.-x/2] dx)) = lim.b[right arrow][infinity](-2[x.sup.2][e.sup.-x/2][|.sub.0.sup.b] - (8x[e.sup.-x/2][|.sub.0.sup.b] + 16[e.sup.-x/2][|.sub.0.sup.b])) = lim.b[right arrow][infinity](-2[b.sup.2][e.sup.-b/2] + 0 - (8b[e.sup.-b/2] - 0 + 16[e.sup.-b/2] - 16[e.sup.0])) = lim.b[right arrow][infinity](-2[b.sup.2][e.sup.-b/2] - -8b[e.sup.-b/2] -16[e.sup.-b/2] + 16) = (0-0-0+16) = 16.

Note that this problem required two applications of integration by parts. Taken directly from Anton's Calculus: A New Horizon 6th ed (1), the following is an example of an integral that can be evaluated using the gamma function (3):

To evaluate [infinity].[integral].0 x[e.sup.-3x] dx using the properties of the gamma function, rewrite the problem as [infinity].[integral].0 [x.sup.2-1][e.sup.-x/(1/3)] dx. Hence, [alpha] = 2 and [beta] = 1/3, and the constant is 1/[GAMMA](2)*(1/3[).sup.2]. Now evaluating the integral gives [infinity].[integral].0 x[e.sup.-3x] dx = [infinity].[integral].0 [x.sup.2-1][e.sup.-x/(1/3)] dx = [GAMMA](2) * (1/3[).sup.2] [[infinity].[integral].0 1/[GAMMA](2) * (1/3[).sup.2] [x.sup.2-1][e.sup.-x/(1/3)] dx] = [GAMMA](2) * (1/3[).sup.2] * [1] = (2 - 1)!(1/9) = 1/9.

Another example of a calculus problem that can be evaluated using the properties of the gamma function comes from Larson, Hostetler, and Edwards's 6th

ed. Calculus (5). Evaluating [infinity].[integral].0 [x.sup.4][e.sup.-x] dx with the properties of the gamma function, gives [infinity].[integral].0 [x.sup.4][e.sup.-x] dx = [infinity].[integral].0 [x.sup.5-1][e.sup.-x/1] dx = [GAMMA](5) * (1[).sup.5] [[infinity].[integral].0 1/[GAMMA](5)*(1[).sup.5] [x.sup.5-1][e.sup.-x/1] dx] = [GAMMA](5) * (1[).sup.5] * [1] = (5 - 1)!(1) = 24

Note that to evaluate this integral using integration by parts would require four applications of the integration by parts method, which would be extremely time-consuming. In the previous examples, the gamma function could be applied directly to the integral, but some integrals can be evaluated by using the gamma function combined with the substitution method. The following is an example out of Anton's Calculus: A New Horizon 6th ed (1).

To evaluate [infinity].[integral].0 [e.sup.-[square root of (x)]]/[square root of (x)] dx with the gamma function, let u = [square root of (x)] and du = 1/2[square root of (x)] dx, now the integral can be written as 2[infinity].[integral].0 [e.sup.-u] du, where [alpha] = 1 and [beta] = 1. Now, evaluating gives 2 [infinity].[integral].0 [e.sup.-u] du = 2 * [GAMMA](1) * (1[).sup.1] [[infinity].[integral].0 1/[GAMMA](1) * (1[).sup.1] [e.sup.-u] du] = 2 * (1 - 1)!(1[).sup.1] * [1] = 2.

Note that 0! = 1. Another problem that the gamma function can be applied to is found in Fisher and Ziebur's 2nd ed. Calculus and Analytic Geometry

(2): Show that for each integer n[greater than or equal to]0, [infinity].[integral].0 [x.sup.n+1][e.sup.-x] dx = (n + 1)[infinity].[integral].0 [x.sup.n][e.sup.-x] dx.

Begin by evaluating the integral on the left hand side of the equation using the properties of the gamma function.

left hand side = [infinity].[integral].0 [x.sup.n + 1][e.sup.-x] dx = [infinity].[integral].0 [x.sup.(n + 2)-1][e.sup.-x] dx = [GAMMA](n+2) * [1.sup.(n+2)] [[infinity].[integral].0 1/[GAMMA](n+2) * [1.sup.(n+2)] [X.sup.(n+2)-1][e.sup.-x] dx] (for [alpha] = n+2, [beta] = 1) = (n+2-1)![1] = (n+1)! = (n+1) * n!

Next evaluate the right hand side of the equation.

right hand side = (n+1)[infinity].[integral].0 [x.sup.n][e.sup.-x] dx = (n+1) [infinity].[integral].0 [x.sup.(n+1)-1] [e.sup.-x] dx = (n+1) * [GAMMA](n+1)*[1.sup.(n+1)] [[infinity].[integral].0 1/[GAMMA](n+1)*[1.sup.(n+1)] [x.sup.(n+1)-1] [e.sup.-x] dx] (for [alpha] = n + 1, [beta] = 1) = (n+1) * (n+1-1)![1] = (n+1) * n!

Since the left hand side equals the right hand side, [infinity].[integral].0 [x.sup.n+1] [e.sup.-x] dx = (n+1) [infinity].[integral].0 [x.sup.n][e.sup.-x] dx, is proven true for all n[greater than or equal to]0.

In addition, the gamma function can be used to solve problems that would be difficult to solve in a reasonable length of time with the integration by parts technique. The following integral is easy to solve when the gamma function is applied; however, it would take 75 applications of integration by parts to solve:

[infinity].[integral].0 [x.sup.75] [e.sup.-x/5] dx = [GAMMA](76) * [5.sup.76] [[infinity].[integral].0 1/[GAMMA](76) * [5.sup.76] [x.sup.76-1] [e.sup.-x/5] dx] = (76-1)!([5.sup.76]) * [1] = 75!([5.sup.76]).

This example demonstrates why the gamma function is such a valuable tool. With the gamma function (3), calculus students can evaluate improper integrals with higher powers and would no longer be limited to problems that only require a few applications of integration by parts. Moreover, since problems in real world applications often cannot be limited merely to problems that can be solved by the integration by parts technique, learning the gamma function in calculus seems especially practical.

Another useful probability density function that can be used in calculus is the beta function, given by f(x) = [GAMMA]([alpha]+[beta])/[GAMMA]([alpha])[GAMMA]([beta]) [X.sup.([alpha]-1)] (1 - x[).sup.([beta]-1)]; where 0 <x<1, [alpha] > 0, [beta] > 0.

Since the beta function is a p.d.f., it integrates to one over its interval, giving 1.[integral].0 [GAMMA]([alpha]+[beta])/[GAMMA]([alpha])[GAMMA]([beta]) [x.sup.([alpha]-1)] (1-x[).sup.([beta]-1)] dx = 1.

The beta function's interval is limited to 0 < x < 1, so it appears somewhat less frequently in calculus books. However, it is still a useful way to evaluate integrals, as shown in the following example taken from Anton's Calculus: A New Horizon 6th ed (1).

To evaluate 1.[integral].0 dx/[square root of (1-x)] with the properties of the beta function, rewrite the integral as 1.[integral].0 [x.sup.1-1](1-x[).sup.1/2-1] dx, showing that [alpha] = 1 and [beta] = 1/2. Thus, the correct constant that makes the integral integrate to one is [GAMMA]([alpha]+[beta])/[GAMMA]([alpha])[GAMMA]([beta]) = [GAMMA](1+1/2)/[GAMMA](1)[GAMMA](1/2). Note that when the constant [GAMMA]([alpha]+[beta])/[GAMMA]([alpha])[GAMMA]([beta]) is introduced, the integral must be multiplied by the constant [GAMMA]([alpha])[GAMMA]([beta])/[GAMMA]([alpha]+[beta]) in order to cancel out what was introduced. Now, evaluating the integral gives 1.[infinity].0 dx/[square root of (1-x)] = [GAMMA](1)[GAMMA](1/2)/[GAMMA](1+1/2)[1.[integral].0 [GAMMA](1+1/2)/[GAMMA](1)[GAMMA](1/2) [x.sup.1-1] (1-x[).sup.1/2-1] dx] = [GAMMA](1)[GAMMA](1/2)/[GAMMA](1+1/2) * [1] = [GAMMA](1)[GAMMA](1/2)/[GAMMA](3/2) = [GAMMA](1)[GAMMA](1/2)/(3/2-1)[GAMMA](3/2-1) = [GAMMA](1)[GAMMA](1/2)/(3/2-1)[GAMMA](1/2) = [GAMMA](1)/(3/2-1) = (1 - 1)!/(1/2) = 0!/(1/2) = 2.

As with the gamma function, the beta function can also be used in application problems, such as the example below, which is an application problem found in Stewart's 4th ed. Calculus (6). If a and b are positive numbers, show that 1.[integral].0 [x.sup.a] (1 - x[).sup.b] dx = 1.[integral].0 [x.sup.b] (1 - x[).sup.a] dx.

Begin by evaluating the integral on the left hand side of the equation using the properties of the beta function.

left hand side = 1.[integral].0 [x.sup.a](1 - x[).sup.b] dx

= 1.[integral].0 [x.sup.(a+1)-1] (1 - x[).sup.(b+1)-1] dx

= [GAMMA](a+1)[GAMMA](b+1)/[GAMMA](a+1+b+1) [1.[integral].0 [GAMMA](a+1+b+1)/[GAMMA](a+1)[GAMMA](b+1) [x.sup.(a+1)-1] (1-x[).sup.(b+1)-1] dx] (for [alpha] = (a+1) and [beta] = (b+1))

= [GAMMA](a+1)[GAMMA](b+1)/[GAMMA](a+1+b+1)[1] = (a+1-1)![GAMMA](b+1-1)!/[GAMMA](a+b+2-1) = a!b!/(a+b+1)!

Next evaluate the right hand side of the equation.

right hand side = 1.[integral].0 [x.sup.b] (1-x[).sup.a] dx

= 1.[integral].0 [x.sup.(b+1)-1] (1-x[).sup.(a+1)-1] dx

= [GAMMA](b+1)[GAMMA](a+1)/[GAMMA](b+1+a+1) [1.[integral].0 [GAMMA](b+1+a+1)/[GAMMA](b+1)[GAMMA](a+1) [x.sup.(b+1)-1] (1-x[).sup.(a+1)-1] dx] (for [alpha] = (b+1) and [beta] = (a+1))

= [GAMMA](b+1)[GAMMA](a+1)/[GAMMA](b+1+a+1)[1] = (b+1-1)![GAMMA](a+1-1)!/[GAMMA](a+b+2-1) = a!b!/(a+b+1)!

Since the left hand side equals the right hand side, 1.[integral].0 [x.sup.a] (1-x[).sup.b] dx = 1.[integral].0 [x.sup.b] (1-x[).sup.a] dx is proven true for all positive numbers a and b. Moreover, the beta function can also be used to solve problems that would be difficult to solve in a reasonable length of time with integration by parts, which is shown in the following example:

1.[integral].0 [x.sup.15] (1-x[).sup.20] dx = 1.[integral].0 [x.sup.16-1] (1-x[).sup.21-1] dx = [GAMMA](16)[GAMMA](21)/[GAMMA](16+21) [1.[integral].0 [GAMMA](16+21)/[GAMMA](16)[GAMMA](21) [x.sup.15] (1-x[).sup.20] dx] = [GAMMA](16)[GAMMA](21)/[GAMMA](16+21) * [1] = (16-1)!(21-1)!/(37-1)! = 15!20!/36! [approximately equal to] 8.6X1[0.sup.-12].

This example shows how the beta function can be used to quickly evaluate certain integrals with large powers, making the beta function an even more valuable tool for integration.

In addition to the gamma and beta functions, another probability density function is found in calculus books. The Cauchy probability density function is given by the function f(x) = 1/[pi]([x.sup.2] + 1); where -[infinity]<x<[infinity].

Since the Cauchy function is a p.d.f., it integrates to one over its entire range of x, giving [infinity].[integral].-[infinity] 1/[pi]([x.sup.2] + 1) dx = 1.

However, since the Cauchy function is an even function, it can also be written as 2[[infinity].[integral].0 1/[pi]([x.sup.2] + 1) dx] = 1 or [infinity].[integral].0 1/[pi]([x.sup.2] + 1) dx = 1/2.

Although calculus books do not call it the Cauchy function, it can be found in nearly every calculus book and is a familiar problem to most calculus students. The following method is the way that calculus books teach students to evaluate the familiar integral [infinity].[integral].-[infinity] dx/(1 + [x.sup.2]), the work is taken directly from Anton's Calculus: A New Horizon 6th ed (1).

[infinity].[integral].-[infinity] dx/(1+[x.sup.2]) = 0.[integral].-[infinity] dx/(1+[x.sup.2]) + [infinity].[integral].0 dx/(1+[x.sup.2] 0.[integral].-[infinity] dx/(1+[x.sup.2]) = lim.l[right arrow]-[infinity] 0.[integral].l dx/(1+[x.sup.2]) = lim.l[right arrow]-[infinity] [ta[n.sup.-1] x[].sub.l.sup.0] = lim.l[right arrow]-[infinity] (ta[n.sup.-1] 0 - ta[n.sup.-1] l [infinity].[integral].0 dx/(1+[x.sup.2]) = lim.l[right arrow][infinity] l.[integral].0 dx/(1+[x.sup.2] = lim.l[right arrow][infinity] [ta[n.sup.-1] x[].sub.0.sup.l] = lim.l[right arrow][infinity] (ta[n.sup.-1] l - ta[n.sup.-1] 0):

Therefore the integral converges and its value is [infinity].[integral].-[infinity] dx/(1+[x.sup.2]) = 0.[integral].-[infinity] dx/(1+[x.sup.2]) + [infinity].[integral].0 dx/(1+[x.sup.2]) = [pi]/2 + [pi]/2 = [pi].

Although there is nothing wrong with that method of solving that integral, it is very time consuming and requires students to go through all of the work every time there is a little variation in the constants of the Cauchy function. A way to cut down on repeated work is to use the properties of the Cauchy p.d.f. to evaluate the integral. For example, to solve the previous integral [infinity].[integral].-[infinity] dx/(1 + [x.sup.2]) using the properties of the Cauchy function, first introduce the correct constant which is 1/[pi]. Note that when the constant 1/[pi] is introduced, the integral must be multiplied by [pi] in order to cancel out what was introduced.

Now, evaluating the integral gives [pi][[infinity].[integral].-[infinity] 1/[pi] 1/(1+[x.sup.2]) dx] = [pi] [1] = [pi].

The following is an example of a similar integral that can be evaluated quickly by using the properties of the Cauchy function. To evaluate [infinity].[integral].-[infinity] 5/(1+[x.sup.2]) dx, first rewrite it in the form of the Cauchy formula by introducing the correct constant 1/[pi]. Now, [infinity].[integral].-[infinity] 5/(1+[x.sup.2]) dx = 5 [infinity].[integral].-[infinity] 1/(1+[x.sup.2]) dx = 5[pi] [[infinity].[integral].-[infinity] 1/[pi] 1/(1+[x.sup.2]) dx] = 5[pi] [1] = 5[pi].

The next example of an integral that can be evaluated with the Cauchy function uses the fact that the Cauchy function is an even function. Evaluating [infinity].[integral].0 8/([x.sup.2]+1) dx is similar to the previous example. First, rewrite it in the form of the Cauchy formula and then introduce the constant 1/[pi], which gives [infinity].[integral].0 8/([x.sup.2]+1) dx = 8 [infinity].[integral].0 1/([x.sup.2]+1) dx = 8[pi] [[infinity].[integral].0 1/[pi] 1/([x.sup.2]+1) dx]. However, since the interval of integration is 0<x<[infinity] instead of -[infinity]<x<[infinity], the get 8[pi] [[infinity].[integral].0 1/[pi] 1/([x.sup.2]+1) dx] = 8[pi][1/2] = 4[pi].

In the previous two examples, the Cauchy function could be applied directly to the integral, but sometimes integrals can be evaluated by using the Cauchy function combined with substitution. Here is an example which is taken out of Larson, Roland, and Hostetler's 5th ed. Calculus of a Single Variable (4).

To evaluate [infinity].[integral].-[infinity] [e.sup.x]/1+[e.sup.2x]dx with the Cauchy function, let u = [e.sup.x] and du = [e.sup.x] dx, now the integral can be written as [infinity].[integral].0 1/1+[u.sup.2] du = [pi] [[infinity].[integral].0 1/[pi] * 1/1+[u.sup.2] du] = [pi] [1/2] = [pi]/2.

From these examples, it is clear that the Cauchy function can be applied to calculus problems, and like the gamma and beta functions, it is also a helpful integration technique.

Finally, another probability density function that can be found in calculus books is the normal distribution. The normal distribution is given by the function

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

The normal distribution integrates to one over its entire range of x, giving

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Like the Cauchy function, the normal distribution is also an even function which means that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

The next example is an integral found in Anton's Calculus: : A New Horizon 6th ed (1), which can be evaluated using the normal function.

To solve [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] using the normal distribution, rewrite it in the form of the normal distribution, [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Note that [mu] = 0 and [sigma] = 1/[square root of (2)].

Now introduce the constant that forces the integral to integrate to one, which is 1/[square root of (2[pi])] * [sigma] = 1/[square root of (2[pi])] * (1/[square root of (2)]) = 1/[square root of ([pi])]. Now multiply the integral by [square root of ([pi])] to cancel out the constant 1/[square root of ([pi])] that was introduced. This gives

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

A similar problem in Larson, Roland, and Hostetler's 5th ed. Calculus of a Single Variable (4) asks students to show that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is convergent, but if the student had already learned about p.d.f.'s, it would be easy to show that the integral converges to [square root of ([pi])]/2. Like the gamma, beta, and Cauchy functions, the normal function can be a useful integration technique in calculus.

CONCLUSION

The research shows that commonly used calculus textbooks do not include a section devoted to probability density functions. Although some calculus books briefly mention the gamma function in an application problem, the usefulness of the function is not explained. Whereas we do not suggest that the p.d.f. method replace the teaching of integration by parts and substitution, we see the need to supplement these conventional techniques with easier methods. This is particularly beneficial to engineering and science majors who are more interested in the application of calculus to their respective disciplines.

In conclusion, I propose undergraduate calculus textbooks include a special section under integration techniques that explains the properties of probability density functions, and demonstrates how to use the properties of the gamma, beta, Cauchy, and normal functions to evaluate integrals. Adding a section about p.d.f.'s in undergraduate calculus textbooks would make certain kinds of integrals easier to evaluate, and reduce the amount of busy work for students. Also, learning the p.d.f.s method in calculus would be helpful to students that continue into statistics and other mathematics courses, such as differential equations.

REFERENCES

(1.) Anton H: "Calculus: A New Horizon, 6th ed." New York: John Wiley & Sons, Inc., 1999.

(2.) Fisher T, and Ziebur L: "Calculus and Analytic Geometry, 2nd ed." Englewood Cliffs: Prentice-Hall, Inc., 1965.

(3.) Hogg R, M. J, and Craig A: "Introduction to Mathematical Statistics, 6th ed." New Jersey: Prentice-Hall, Inc., 2005.

(4.) Larson R, Hostetler R, and Edwards B: "Calculus of a Single Variable, 5th ed." Lexington: D.C. Heath and Co., 1994.

(5.) Larson R, Hostetler R, and Edwards B: "Calculus, 6th ed." Boston: Houghton Mifflin Co., 1998.

(6.) Stewart J: "Calculus, 4th ed." Pacific Grove: Brooks/Cole Publishing Co., 1999.

Stacy Pitman

Andreas Lazari

Department of Mathematics and Computer Science

Valdosta State University

Valdosta, GA 31698

(229) 333-5778

Corresponding author: Andreas Lazari

E-mail: alazari@valdosta.edu

(229) 333-7154