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The recipe is quite simple to understand looking at an example (and it is understood best by having paper and pencil to follow it): f:= x -> x^2 the parabola with its inverse g:= y -> sqrt(y). Say you want the integral of g over 0 ... 2, which (here) is the area between the graph and its horizontal axis. That is the same as the area of the rectangle minus the area between the graph of g and the vertical axis, where the rectangle has corners 0, 2 and g(0)= f^(-1)(0) and g(2)= f^(-1)(2). Now recall the geometric interpretation of the compositional inverse of a function: it is reflection at the diagonal. The reflection does not change the rectangle's area. However it maps the area between graph and vertical axis to the area between the the reflected graph and the reflected axis, which now is the horizontal. But that is the integral of f over f^(-1)(0) ... f^(-1)(2). Thus we have: 
  Int(g(y), y=0...2) = areaRectangle - Int('f'(x), x=0...sqrt(2)).
In Maple's language: 
  eta0:=0; eta1:=2;  
  xi0:='g(eta0)'; xi1:='g(eta1)';

  areaRectangle:= '(eta1-eta0)*(xi1-xi0)';

  'int(g(y), y= eta0 .. eta1)': '%'=%;
  'areaRectangle' - 'int(f(x), x= xi0 .. xi1)': '%'=%;

                          eta1
                         /                  1/2
                        |                4 2
                        |      g(y) dy = ------
                        |                  3
                       /
                         eta0


                                  xi1
                                 /                 1/2
                                |               4 2
               areaRectangle -  |     f(x) dx = ------
                                |                 3
                               /
                                 xi0
Thus you only invert at the boundary and do not need the expensive (and having possible rounding errors) inverse for integration. In good cases Maple even knows a symbolic integral for f, while it may not have a closed form for g and everything would have have to be done in a costly numerical setting otherwise. To sketch the proof: more formally use 
  Int( g(y), y ) = Int ( x*D(f)(x), x ) # change y=f(x), g°f = id
                                        #
  = x*f(x) - Int( f(x), x)              # integrating by parts
and apply the fundamental theorem of calculus. In Maple for F and G being inverse to each other: 
  with(IntegrationTools):

  Int( G(y), y )= ``;
  ``=Change(lhs(%), y=F(x), x): convert(%,D);
  ``=subs(G(F(x)) = x,rhs(%));
  ``=Parts(rhs(%),x);

                             /
                            |
                            |  G(y) dy =
                            |
                           /


                            /
                           |
                        =  |  G(F(x)) D(F)(x) dx
                           |
                          /


                               /
                              |
                           =  |  x D(F)(x) dx
                              |
                             /


                                      /
                                     |
                         = F(x) x -  |  F(x) dx
                                     |
                                    /
Actually one simply computes 
                eta1                    xi1
               /                       /
              |        (-1)           |
              |      (f    )(y) dy =  |     x D(f)(x) dx
              |                       |
             /                       /
               eta0                    xi0


                                              xi1
                                             /
                                            |
               = f(xi1) xi1 - f(xi0) xi0 -  |     f(x) dx
                                            |
                                           /
                                             xi0
The appended sheet gives a more detailled motivation. Download 102_integrate_inverse_function.mws
View file details
example convert calculus integral

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