Module 9 : Integral Calculus

902 : Interesting Integrals

P U R P O S E

The purpose of this project is to learn how to perform definite and indefinite integration using Maple, and to explore various properties and theorems of integrals.

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A. Definite and Indefinite Integral

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We define a function to use as a guinea pig.

> f := x -> .01*x^4 + 3 + sqrt(x) + 6/x^2 + x*sin(3*x);

and graph it while marking off an area to compute

> plot( { f(x), [[1,0],[1,f(1)]], [[6,0],[6,f(6)]] }, x = 0..7, y = 5..20);

With these commands we compute the indefinite integral. Note that the Int command only displays the integral without computing its value, while the int command computes the anti derivative.

> Int( f(x), x ); value(%); int( f(x), x );

We can also compute the definite integral which represents area under the curve. Again, the Int command just sets it up, and the int command gets the numerical answer.

> Int( f(x), x = 1..6); value(%);
int( f(x), x = 1..6);

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B. Linearity Properties

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A process that takes T(aáf(x)) + bág(x)) to aáT(f(x)) + báT(g(x)) is called a linear transformation. The derivative is an example of a linear transformation. The integral is another example.

Lets define some functions, and explore the proposition that the integral of a sum is the sum of integrals :

> f := x -> x^3 + 10*x + 3;

> g := x -> sin(10*x);

> Int( c*f(x), x = 0..14) : % = value(%);

> c*Int( f(x), x = 0..14) : % = value(%);

The fact that the results are the same is an indication (but does not formally prove) that this proposition is valid.

Lets explore the proposition

> Int( f(x) + g(x), x) : % = value(%);

> Int( f(x), x ) + Int( g(x), x ) : % = value(%);

The fact that the results are the same is an indication (but does not formally prove) that this proposition is valid.

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C. Upper and Lower Bounds

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An upper bound for a function is a constant M which the function never exceeds in a given interval : f(x) M. Similarly, a lower bound, m, for a function is a value which is never larger than the value of the function : m f(x) over some interval. There are many functions which are difficult or impossible to integrate. However, using the upper and lower bound of a function, we can get upper and lower bounds for an integral :



In the student package of commands, there are maximize and minimize commands which find the maximum and minimum value of a function over an interval.

> restart; with(student):

> g := x -> 1/3*x^3 - 7*x^2 + 35*x + 30;

> minimize( g(x), x=0..14); m := evalf(%);
maximize( g(x),x=0..14); M := evalf(%);

Here is snapshot of the situation.

> plot( { g(x), m, M}, x=0..14, y=0..90 );

The horizontal lines indicate the minimum and maximum values. The value m(b-a) represents the area of the smaller rectangle which fits under the curve, and the value M(b-a) represents the larger rectangle which the curves fits within.

Lets now verify that

> Int( m, x=0..14): % = value(%);
Int( g(x),x=0..14): % = evalf(value(%));
Int( M, x = 0..14): % = value(%);

Here you can see the min and max. Thus we have that m g(x) M for the interval. What can we say about the integral of g(x) if were not able to compute it directly?

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D. The Absolute Value and Integrals

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Lets explore the proposition that and see why its true.

Let define a function as an example, and then compute each integral and see which one is larger.

> f := x -> x^3 - 10*x^2 + 20*x + 10;

Here is

> Int( abs( f(x) ), x = -3..8): % = evalf( value(%));

and

> abs( Int( f(x) , x = -3..8)): % = evalf( value(%));

The answers are different. If we look at the plot of each function we can see some clues as to why there is a difference.

> plot( f(x), x = -3..8, y = -15..25);

> plot( abs(f(x)), x = -3..8, y = -15..25);

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E. The fundamental Theorem

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One version of the Fundamental Theorem is that if , then FÕ(x) = f(x). We begin be defining f(x), then define F(x) by way of an integral of f(x).

> f := x -> x^2 - 10*x + 30;

> F := x -> int( f(t), t = 0..x);

Locate f(x) by noting the Y-intercept fo 30. The other graph is F(x) which is measuring the area under f(x).

> plot( { f(x), F(x) }, x = 0..10);

Note that if we compute the derivative of F(x) we get f(x) as we should.

> F(x); diff( F(x), x);

>