1. Taking Antiderivatives

We have an inert version "Int" and the version "int" which actually calculates.

>	exint := Int((x^2-1)/(x^5+1),x);

>	value(exint);

>	anti_derivative := int((x^2-1)/(x^5+1),x);

The second time, the answer came much faster, although there seems to be no remember table:

>	interface(verboseproc=3);

>	op(4,eval(int));

>	print(int);

>	op(4,eval(`int/int`));

So getting to the remember table is a bit more involved...

The second thing we may want to do is to verify the answer.  We can take the derivative of the recently computed antiderivative.

>	da := diff(anti_derivative,x);

Why does this seem so different from out original (x^2-1)/(x^5-1)?????

>	testeq(da=(x^2-1)/(x^5+1));

The probability one test, built-in in the "testeq" command returns inconclusive.

>	f := unapply(da,x):

>	g := x -> (x^2-1)/(x^5+1):

>	r := evalf(rand()/10^12);

>	evalf(f(r) - g(r));

So there is not really a difference.

>	pf := plot(f(x),x=0..100): pg := plot(g(x),x=0..100):

>	plots[display](pf,pg);

2. Definite Integration

Let us move to definite integration, and verify the fundamental theorem of calculus.  We can compute definite integrals via antiderivatives.

>	f := 1/x^2;

Let us compute an antiderivative:

>	ad := int(f,x);

To compute the integral of f from a to b, we apply the fundamental theorem of calculus:

>	symbolic_integral := subs(x=b,ad) - subs(x=a,ad);

For example:

>	subs(a=-1,b=1,symbolic_integral);

However:

>	numerical_integral := int(f,x=-1..1);

The numerical_integral is right:

>	plot(f,x=-1..1,view=[-1..1,0..100]);

>

3. Numerical Integration

In many cases, there is no symbolic antiderivative.

>	integrand := exp(cos(x));

>	int(integrand,x);

What we still can compute are definite integrals, but then numerically:

>	integral := Int(integrand,x=0..1);

>	val := evalf(integral);

Here we see the usefulness of the inert Int command, because it does not make sense to first try a symbolic antiderivative.

>	macro(numint = evalf@Int);

>	numint(integrand,x=0..1);

A direct numerical integration routine was invoked, without first trying to integrate it symbolically.

4. Integral Transforms

To go from the time domain to the frequency domain, we apply integral transforms.

>	diffeq := diff(y(t),t$2) + y(t) = sin(2*t);

>	inits := y(0) = 2,D(y)(0) = 1; # initial conditions

>	inttrans[laplace](diffeq,t,s);

>	subs(laplace(y(t),t,s)=Y(s),%);

>	subs(inits,%);

We can solve this equation algebraically, find Y(s) and then go back to the time domain via invlaplace.

5. Integrals with parameters

Maple 7 would have evaluated the following integral:

>	int(f,x=a..b);

>

which would be wrong if a*b < 0.

>	int(f,x=a..b) assuming a>0,b>0;

Equivalently, we can impose a global assumptions on a and b:

>	assume(a>0,b>0);

>	int(f,x=a..b);

The value of the integral depends on whether a and b have the same sign.

6. Summation

>	sum('i','i'=0..n);

>

>