Page 1 of 1

A Calculus Primer Part IV- Introduction to Integrals

#1 macosxnerd101  Icon User is offline

  • Self-Trained Economist
  • member icon




Reputation: 10180
  • View blog
  • Posts: 37,585
  • Joined: 27-December 08

Posted 01 May 2011 - 03:20 PM

Integral calculus deals with accumulations over an interval. We can also call this a summation, which leads us to deal with series. Integrals are also called anti-derivatives based on the fundamental idea of calculus that we add up a rate to get a change.

Let's start looking at the series definition of an integral:
Attached File  summation_limit.bmp (12.77K)
Number of downloads: 257

As we see, we have an infinity series, with a function f(x), as well as a deltaX term. The deltaX term is equivelant to (b-a)/n, where a and b are the limits of integration, or the points defining the interval being integrated over. As n grows without bounds, the number of partitions of the function (deltaX) becomes infinite as well. Thus, this provides the area under the curve of the function.

Now for our integration function: Integral(f(x) dx). Here, our dx, also called a differential or infinitesimal, tells us a few things. First, it is our partition of f(x), like the deltaX in the above limit definition. Remember, integrals are used to add up a lot of individual partitions. Second, the differential tells us the variable to integrate over. This becomes more important when dealing with functions with multiple variables and differential equations.

Now let's take a look at our rules for integration. Looking back at my tutorial on derivatives, we see our rules for derivatives. To integrate, we start on the right-hand side of our equations below, and end up with the terms on the left-hand side. So for example, Integral(cos(x) dx) = sin(x). Notice how the derivative of sin(x) = cos(x).

Quote

d/dx k = 0
d/dx x = 1
d/dx xn = n * xn-1
d/dx (f(x) + g(x)) = d/dx f(x) + d/dx g(x)
d/dx kf(x) = k * d/dx f(x)
d/dx sin(x) = cos(x)
d/dx cos(x) = -sin(x)
d/dx tan(x) = sec^2(x)
d/dx sec(x) = sec(x)tan(x)
d/dx csc(x) = -csc(x)cot(x)
d/dx cot(x) = -csc^2(x)
d/dx ln(x) = 1/x
d/dx logb(x) = 1/(x ln(B))
d/dx e^x = e^x
d/dx a^x = ln(a) * a^x
d/dx |u| = du/dx when to the right of the vertex, -du/dx when to the left of the vertex, and undefined at the vertex


There are a couple of these rules that are slightly different for integration than the derivative form. First, the Integral(du) = u. This goes back to the idea that if we add up a rate (du), we get a change/amount (u). The second integral that is different than in the above table is Integral(1/x dx) = ln|x| rather than ln(x) because ln(x) is bounded in the first quadrant for all x >= 1.

Notation
In Calculus, it is common to refer to F(x) as the function representing Integral(f(x) dx). So here, F'(x) = f(x).

Indefinite Integrals
The above integrals we have seen thusfar simply demonstrate the concept of anti-differentiation. There are a couple more things to consider with integrals. Our first type of integral is an indefinite integral. This means we are integrating a function without limits of integration. This does not tell us the area under the curve. Rather, it returns a family of functions such that, given one of those functions and two points a and b such that a <= b, F(b)- F(a) is the area under the curve.

So how do we evaluate our indefinite integrals? The Integral(f(x) dx) is evaluated by going backwards from our derivative rules above, based on the terms for f(x). A constant + c is then added to the end of the returned function. The reason for adding this constant is because we are dealing with a family of functions. Let's think about this a different way. An integral is an anti-derivative, so it makes sense that differentaiting F(x) will return the original function f(x) (or in other words, as was mentioned above, F'(x) = f(x)). Since constants d/dx constant = 0, the added constant when integrating accounts for the fact that the original function that provided that derivative may have originated elsewhere.

Let's look at a couple of examples:

1) F(x) = Integral(sin(x)dx) = -cos(x) + c
Based on our derivative rule that d/dx cos(x) = -sin(x), we know that the Integral(sin(x)dx) has to have a cos(x) term in it. The reason there is a -1 coefficient for cos(x) is to cancel out the other -1 when differentiating cos(x). So d/dx(-cos(x) + c) = sin(x).

2) F(x) = Integral(e^x dx) = e^x + c
This is perhaps one of our easieset integrals. If we differentiate F(x), notice how we end up at e^x.

3) F(x) = Integral(3x^2 dx) = x^3 + c
For this integral, the coefficient of 3 can be pulled in front of the integral, in the same way that it could with a derivative. Remember, integrals and derivatives are simply limits, so the same rules taught in precalculus for limits apply here as well. So for the Integral(x^2 dx), simply use the power rule. Add one to the exponent, then divide by that new exponent. So x^3/3 * 3 + c = x^3 + c.

Definite Integrals
A definite integral represents the area under the curve of a function between two points a and b. We evaluate Integral(a, b, f(x), dx). The same integral rules apply; however, there are two differences. First, the addition of a constant is not included. Thinking about this more, there is the same area under the curve for all functions of F(x), regardless of where they are. More analytically, the constants simply cancel out when evaluating the area. Another way of thinking of definite integrals is the non-overlapping area of Integral(0, b, f(x) dx) and Integral(0, a, f(x), dx).

Let's work through some examples:

1) Integral(0, 5, 3x^3 + 4x^2, dx)
= ((3/4 * x^4) + (4/3 * x^3)) |(x = 0, 5)
= (1875/4 + 500/3)

Since the first limit of integration is x = 0, there is nothing to subtract from the first term.


2) Integral(-pi/2, pi/2, sec^2(x) dx)
= tan(x) |(x = -pi/2, pi/2)
= tan(pi/2) - tan(-pi/2)
= 0

For this integral, the answer is 0. Looking at the unit circle, the domain of tan(x) is -pi/2 to pi/2, so it is defined in the first and fourth quadrants. Since these quadrants are symmetrical, and one is above the x-axis and the other is below it, the areas cancel out.


Fundamental Theorem of Calculus
Throughout this tutorial, the fundamental theorem of calculus has been used many times. It also serves to relate differential and integral calculi. This last section will simply serve to put the name with the concept. There are two parts to the fundamental theorem of calculus:
  • F(x) = Integral(0, x, f(t) dt)
  • Integral(a, b, f(x) dx) = F(b) - F(a)


The first part of the fundamental theorem of calculus goes back to the idea of adding up a rate to get a change. In other words, the sum of all the changes from 0 to x is the net change at that point. From this, it can also be inferred that d/dx Integral(0, x, f(t) dt) = f(x) * x'. Or in other words, F'(x) = f(x), as was stated above.

Let's look at an example:
d/dx Integral(0, 2x, f(x) dx) = 2f(x)

Based on the fundamental theorem of calculus, d/dx F(x) = f(x). The reason the answer has a coefficient of two is due to the chain rule when differentiating. Using the second part of the fundamental theorem, the answer comes out to d/dxF(2x) - F(0). When differentiating F(2x), it comes out to 2*f(x).


Conclusion
This tutorial has introduced the basics of integration, as well as the concepts behind it. My next tutorial will discuss more techniques for integration, and the following tutorial will cover applications of integrals.

Is This A Good Question/Topic? 2
  • +

Page 1 of 1