Integration

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Integrals:

The process of finding derivatives and using them to solve problems is called differentiation. Think about derivatives as analyizing a problem at various instants, since it will describe the parent function at particular points of interest. The process of Integration does exactly the opposite. It will take instants and bring them together as a whole, so to speak. When dealing with integration, we generally know the derivatives and seek the parent function. So implied in the name of the process. By going from instants to the whole gives us a path to an unknown parent function that we seek in our problem. Generally speaking, most students find this reverse process much more difficult since it involves a indept knowlege of derivative properties and more precise decision making for success. I will give you a very brief and informal description on how this works and give some basic examples of a few types of problems that uses integation to solve them. The integration process is also referred to as anti-differention.

The integral(or anti-derivative) of a function, f(x), is symbolized by ∫f(x)dx. The differential dx indicates the variable of integral is always written to the far right. We are seeking a parent function with a derivative of f(x).

The expression between the integral sign and the differential is the expression to be integrated. This representation is for a function of a single variable, usually studied in Calc I & II.

In multivariable calculus (calc III), the function to be integrated is usually dependent on several variables, i.e., f(x,y) or f(x,y,z) and multiple integrals appear. For example, ∫∫f(x,y)dxdy and ∫∫∫f(x,y,z)dxdydz represent a double and triple integral in calculus III and can be done by two and three separate integrations, respectively. They are done by separate integrations starting from the inside out. For example, in the case of the triple integral, the first integration is done with respect to x, then with respect to y, and finally with respect to z. The other variables in the expression f(x,y,z) are treated as constants when performing the integration with respect to one of them. For multiple integrals, the order of integrations is always done from the inner most variable out to the variable on the extreme right. (These variables are indicated by their differentials). Don't worry about these. Only those math majors who survive Calculus I & II will get the opportunity to be exposed to them.

These are examples of Indefinite Integrals, since no limits (numbers or symbols that appear at the bottom and top of the integal symbol) appear on the integration symbol. So, the results will be variable expressions with unknown constants (assuming no conditions on the problem are given). Definite Integrals have limits on the integration symbol (lower & higher).

So, going from a derivative to its parent function will involve integrating the derivative. However, for indefinite integrals, the result will contain many answers since an unknown constant will be present. In symbols, ∫f(x)dx=F(x) + C. I use F(x) in the result since that function will differ from f(x). The constant C is arbitrary and could take on an infinite number of values, if no other information is given in the problem.

It will always be the case that F'(x) = f(x). This is a fundamental relationship that ties both processes together. This is one of the fundamental theorems of calculus. For example, ∫2xdx=x² + C. Note that the derivative of x^2 is 2x, which is the expression in the integral. However, only the simplest of integrals can be done by guess work. Rules are established for integration so more complicated integrals can be evaluated. There are expressions that are very complicated & special techniques are studied (calc II) and other integrals that can’t be solved. However, most applications deal with definite integrals. Their values can be approximated very closely. The calculator will do this for us very quickly (the TI-83 calculator).

Some Integration formulas: There are numerous integration forms that are listed in many books. It's virtually impossible to know them all. I will list just a few popular ones that seem to be used in the most basic of problems. What makes this difficult is that the student needs to pick the correct form for their integral. Not a very easy task. If the wrong form is chosen, the problem will not be able to be solved. The following are a few of the basic rules.




Basic integration formulas with examples

1) ∫du = u + c
EXAMPLE: ∫dx = x + c

2) ∫k f(u)du = k∫f(u)du (k any constant)
EXAMPLE: ∫2pxdx = 2p∫xdx

3) ∫[f(u) + g(u)]du = ∫f(u)du + ∫g(u)du (integrate a sum over each term)
EXAMPLE: ∫(x² + 2x + 3)dx = ∫x2dx + ∫2xdx + ∫3dx= ∫x2dx + 2∫xdx + 3∫dx

4) (POWER RULE): ∫u⁽ⁿ⁾du = u⁽ⁿ⁺¹⁾ / (n+1) + C (Must make sure du is present)
EXAMPLE: ∫x²dx = x³ / 3 +C, since u=x, n=2, du=dx & is present.
EXAMPLE: ∫(5x²+ 7)4xdx, since u=5x²+7, du=10xdx (need to multiply inside by 10 & compensate by dividing on the outside by 10)
(1/10)∫(5x²+7)⁽⁴⁾10xdx = (1/10)[(5x²+7)⁵]/5 + C = (5x²+7)⁵/50 + C

5) ∫e^udu = e^u+ C
EXAMPLE: ∫e^5xdx = (1/5)∫e>^5x5dx = (1/5)e^5x+ C

6) ∫du/u = ln |u| + C
EXAMPLE: ∫xdx/(x² +7) = 1/2 ∫2xdx(x²+7) = (1/2)ln(x²+7)+C
EXAMPLE: ∫(sin3x)dx/cos3x = (-1/3)∫(-sin3x)3dx/cos3x =(-1/3)ln |cos3x|+C

7) ∫cos(u)du = sin(u) +c
EXAMPLE: ∫ {[cos(lnx)] /x} dx = ∫ [cos(lnx)] [1/x] dx = sin(lnx) + C

8) ∫sin(u)du = -cos(u) +C

9) ∫sec²udu = tan(u) +C

10) ∫ sec(u)tan(u)du = sec(u) +C
EXAMPLE: ∫xsec(x²)tan(x²)dx = (1/2)∫ sec(x²)tan(x²) 2xdx= (1/2)sec(x²) + C

11) ∫ csc(u)cot(u)du = -csc(u) + C

12) ∫ csc²udu = -cot(u)+C

~NOTE: An integral without an differential is meaningless.

In multivariable calculus, derivatives & integrals are taken with respect several different variables. Partial derivatives & multiple integrals are used. So, when differentiating, one must state reference to the variable. (on occasions, in single varaible calculus, it is understood).

However, not always. For example, the derivative of x wrt to x is 1, but the derivative of x wrt to t is dx/dt. Likewise for integration. The differential to the far right of an integration gives the variable of integration. In multiple integrations (double & triple integrals), several different differentials are present. So, one would integrate wrt one, then integrate again, wrt another, & so on. These techniques are performed by treating all other varaibles as constants (temporally) until the operations are completed. Don't be concerned with this unless you find yourself in a Calculus III course.


THE DEFINITE INTEGRAL

Question: "what is a definite integral?"
Answer: "A number" (that's it!)

Question: "what does it represent?"
Answer: "Possibly an area, volume, length, or some other numerical value of some quantity."

Question: "can it be a negative number?"
Answer:"sure can"

To understand how this can happen, we go to the basic area problem.

Some simple areas can be found by inscribing or circumscribing rectangles between the curve and the x axis and then letting the number increase to infinity. If the sum is formulated properly, the limit of the sum will equal the area in question. Be careful, since sums below the x axis are negative.

These sums are called Riemann Sums.

The tie between the limit of a Reimann Sum and the area in question can be shown in a very involved proof. It is lengthy and requires much background in calculus. It is omitted here.

It also can be proven, in a less involved way, that the area under a curve and between x=a and x=b can be found by integration. (uses one of the fundamental theorems, namely, the tie between differentialion and integration) . All you have to do is to set up the definite integral from a to b and evaluate it.

The definite integral from x=a to x=b is denoted as ∫_a^bf(x)dx and is evaluated by integrating f(x), then substituing the upper limit in the expression, then subtracting the lower limit substituted in the expression. If we denote F(x) as the integral (without the constant), it would look this way: F(b) - F(a).

~Well, since the limit of a Riemann Sum equals the area and a definite integral equals the area, we define the definite integral as the limit of a Riemann Sum. Therefore, many problems are solved by setting up Riemann Sums.

~Consequently, defining the definite integral as this limit, we have a direct tie to the area problem and other problems that can be expressed and formulated as limits of these types of Reimann Sums. This will give us a way to solve all kinds of applied problems (more complicated area problems, volume problems, surface area problems, work problems, arc length problems, fluid pressure problems, just to name a few).

~Therefore, if a problem is subject to analysis by using a limit of a Reimann Sum, it can (theoretically) be solved by a definite integral.

~Basically, Here is how the problem must be formulated

lim ∑f(kx)Δx= ∫_a^bf(x)dx
Δx→ 0

So, informally, if a problem involves a solution from x=a to x=b, we partition the interval into k subintervals, each of length Δx, form the Riemann sum, then set up it's limit. In actuality, only one representative term of the sum is needed since they are all formed the same way (this is known as a "specimen" piece).(i.e., to find the volume of a solid, we can think of the solid as the sum of an infinite number of smooth disks of different sizes. However, the volume of each one is represented the same way, so all we need is a specimen disk in the formulation that represents each one. This would be called our specimen volume piece) However, we don't need to evaluate the limit, just set it up so that we can recognize the funcion f. Then all we have to do is integrate f(x) from a to b for our answer. Solid disks may not be appropiate for all volumes. Depending on their cross-section and how they were generated, a different specimen piece must be chosen (i.e., washers, shells, or a special one with a given cross-sectional area to fit our problem).

~Informally, one can think of the symbol ∫_a^bf(x)dx as the lim∑over [a;b] and the Δx in the sum becoming dx in the integral.

~It turns out that many applied problems can be "set up" by using Riemann Sums of these types, therefore, can be solved by a definite integral.

~For those of you who go on to calculus II, you will see many.

~The above mention problems are at the heart of calculus II.

Thank goodness for our TI-83 calculator since the number associated with any definite integral can be found easily. You do not need to know any integration rules. It will be done automatically. An added feature along with the answer will be a sketch of the curve inticating an area that the value of the definite integral represents. Very, very, Nice!