Reduction of integration
- From: "Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx>
- Date: Sun, 22 Oct 2006 03:23:24 GMT
Take an example:
Int(x^alpha,x=a..b)
choose x = u^beta ==> dx = beta*u^(beta - 1)du
on substitution we have
x^alpha * dx = u^(alpha*beta) * beta * u^(beta - 1) * du =
beta*u^(beta*(alpha + 1) - 1) * du
Since beta is a free parameter we can choose it so that the exponent above
is 0 which implies that beta = 1/(alpha + 1)
hence
Int(x^alpha,x=a..b) = Int(1/(alpha + 1),u=a^(alpha + 1)..b^(alpha + 1))
the second integral is easily caried out to be 1/(alpha + 1)*[b^(alpha +
1) - a^(alpha + 1)]
This can be found by directly but it seems to maybe offer some some
potential? In the above derivation I only use simple algebra and
differentiation and no integration except at the final step which was ver
simple. I'm wondering if such techiniques are widely used? It would see
that it might very hard to find a family of substitutions to reduce the
integrand but I'm wondering if its always possible? (since it only involves
taking derivatives)
i.e., if in general I have Int(f(x),x=a..b) can I find a "family" of
functions x = g(y) such that a specific member of the family of the form
f(g(x))*g'(x)dx is simple to integrate? It would seem that potentially this
is only true for functions that have inverses and one is essentially
plugging in the inverse to reduce the integrand. Its not quite the inverse
though and maybe it works for functions that are not invertable?
Is there any area of mathematics that deals with the questions I'm asking?
Is this a trivial problem? Its obviously one of integrand transformations
but by adding a parameter(or maybe even more) it seems like one might be
able to simplify some more complicated integrals. I think one has to ask
the question does there exist a function x = g(y) such that on substitution
into f(x)*dx we get something much simpiler to integrate.
Am I making a big deal out of nothing?
Thanks,
Jon
.
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