[Maxima] Discontinuity Points

Aleksas Domarkas aleksasd873 at gmail.com
Wed Jun 27 00:34:47 CDT 2012

```  How to find the points of discontinuity af a function on interval?

Application
Theorem(Generalized Newton-Leibnitz formula).
Let F = integrate(f,x) and
x[1], x[2], ..., x[n]  discontinuity  points of  F  from interval [a, b],
p is list:  p=[a, x[1], x[2], ..., x[n], b]. Then
integrate(f,x,a,b)=sum(F(p[k+1]-0)-F(p[k]+0),k,1,n+1)

Example.  integate(cos(3*x)/(5-4*cos(x)),x, 0, 10*%pi)
(%i1) f:cos(3*x)/(5-4*cos(x))\$
(%i2) F:integrate(f,x)\$
discontinuity  points of  F is    %pi, 3*%pi, 5*%pi, 7*%pi, 9*%pi
(%i3) p:[0,%pi,3*%pi,5*%pi,7*%pi,9*%pi,10*%pi];
(%o3) [0,%pi,3*%pi,5*%pi,7*%pi,9*%pi,10*%pi]
(%i4)
'integrate(f,x,0,10*%pi)=sum(limit(F,x,p[k+1],minus)-limit(F,x,p[k],plus),k,1,6);
(%o4) integrate(cos(3*x)/(5-4*cos(x)),x,0,10*%pi)=(5*%pi)/12

Wrong:
(%i5) integrate(f,x,0,10*%pi);
(%o5) 0
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