Improper Integrals

> restart:libname:=`C:\\PROGRAMS\\MAPLEV/lib`,`C:\\PROGRAMS\\MAPLEV/lib/CalcP`:with(CalcP):

Note: The F.T.C. says that where F is any antiderivative of f. It follows from this that = F(1) - F(-1) = -2 where . Does the plot of below show this as a sensible answer?

> plot(1/x^2,x=-4..4,y=0..8,discont=true,colour=black);

Recall: The F.T.C. says that only if a and b are numbers and f is continuous on [a,b].

Type 1. Integrals over infinite intervals

e.g. Evaluate .

> f:=x->1/x^2:plot(f(x),x=0..10,y=0..1,discont=true,colour=black,title=`y = 1/x^2`);

Define F(b) = the area under y = f(x) from x = 1 to x = b:

> F:=b->Int(1/x^2,x=1..b);

> some_areas:=value([F(1),F(2),F(3),F(4),F(10),F(100),F(10000),F(b)]);

Clearly, the bigger b gets, the closer gets to 1. It seems natural to define to be the limit as b -> . That is: = .

Definitions

a) If exists for every t bigger than or equal to a then if the limit exists.

b) If exists for every t less than or equal to b then if the limit exists.

If the limits exist then the integrals are convergent. Otherwise, they are divergent .

c) if both the intgrals are convergent .

e.g.s

> assume(b<-1);f:=int(1/x,x=b..-1);Limit(f,b=-infinity)=limit(f,b=-infinity);Int(1/x,x=-infinity..-1)=int(1/x,x=-infinity..-1);

> f:=x*exp(1)^(-x^2);

> plot(f,x=-4..4);

> F:=int(f,x);

> leftside:=int(f,x=-infinity..0);rightside:=int(f,x=0..infinity);

> Int(f,x=-infinity..infinity)=int(f,x=-infinity..infinity);

Determine p so that is convergent.

> f:=1/(x^p);

> Int(f,x=1..t):%=value(%);

> f1:=simplify(%);

The results are easier to see if you write the above expression as

Fact: converges if p > 1 and diverges if . Now, the hard part, prove it!

Show that Gabriel's Horn can be filled with paint but not painted!

> `f:=1/x;

> revolve(f,x=1..10,nocap);

1. Find the volume of Gabriel's Horn:

> V:=Int(Pi*(1/x)^2,x=1..infinity);volume:=value(V)*`cubic units`;

Note: part 1 "says" that we can fill Gabe's horn with about 3 cubic meters of paint.

2. Find the surface area of Gabriel's Horn.

In order to do part 2 we will need the following goft from the God of Calculus: If f is positive on [a,b] and has a continuous derivative on [a,b] then the surface area of the surface obtained by rotating between a and b about the x-axis is .

> g:=2*Pi*f*sqrt(1+diff(f,x)^2);

> S:=Int(g,x=1..infinity);surface:=value(S);

Go figure!!

Type 2. Integrals with discontinuous integrands.

e.g. Find

> f:=1/surd((x-1),3):plot(f,x=-1..2,y=-5..5,discont=true,colour=black,title=`y = 1/(x-1)^(1/3)`);

Note that the function is discontinuous at x = 1 so the FTC does not apply .

Let's experiment:

> A:=t->Int(f,x=t..2);

> value([A(1.5),A(1.1),A(1.01),A(1.001),A(1.0001),A(1.00000001)]);

=> it looks like is getting closer and closer to 1.5 and it seems like we should define = (and we will!).

Definitions:

a) If f is continuous on [a,b) and discontinuous at b then if the limit exists.

b) If f is continuous on (a,b] and discontinuous at a then if the limit exists.

If the limits above exist the integrals are convergent . Otherwise they are divergent.

c) If f is discontinuous at c, a < c < b, then if both the integrals converge .

e.g.s

1.

> f:=1/x^2;Int(f,x=-1..0)=int(f,x=-1..0);

2.

> g:=2/(x^2-2*x);convert(g,parfrac,x);plot(g,x=-4..4,y=-4..4,discont=true,colour=black);I1:=Int(g,x=1..2)=int(g,x=1..2);

3. N.B. There are two types of improper integrals in the next one!!

> f:=1/(sqrt(x)*(1+x));plot(f,x=0..10,colour=blue,thickness=2);

> Int(f,x=0..1)+Int(f,x=1..infinity)=int(f,x=0..1)+int(f,x=1..infinity);