Unformatted text preview: val [0, +1] is NOT FINITE. From a Geometric point of view we may define
INTEGRABLE as :
f(x) is INTEGRABLE ⇔ the area under f(x) is FINITE
175 ∞ The discrete summation
discrete summation
discr 1 Σ /2 i CONVERGES. The terms are finite in value.
i=0 However, the subscript i varies over the infinite interval [0, +∞).
∞ −i
also CONVERGES.
Likewise the discrete summation i Σ0 e
discrete
=
It is possible to have an integral or a continuous summaiton over an infinite
continuous
interval.
+∞ It is now easy to see that the continuous summation ∫
continuous
0 {e−x} dx must CONVERGE even though the interval [0, +∞) is infinite. Hence f (x) = e− x is INTEGRABLE. +1 f (x) = e− x x 0 +∞ ∫ 0 {e−x} dx +∞
−x ]
= [− e
0 = (− e −∞) − (− e −0 ) = 0 − (− 1) = +1 176 We may also integrate over (−∞, +∞) as in the example below. Let
f(x) = { e− x for x ≥ 0
e x for x ≤ 0 +1
e− x ex 0 −x
+∞ 0 x
+∞ ∫ f(x) dx = ∫ f(x) d x + ∫ −∞ −∞ 0 f(x) dx = +1 +1 = +2 associati
We are familiar with the associativity property in addition : (a+b)+c = a+(b+c).
associativity
This property holds good when the number of terms are finite. It also holds good for
a discrete summation of COUNTABLY many terms if all the terms are of same
discrete summation
discr
sign. But it breaks down when there are COUNTABLY many terms of both positive and
negative sign. The alter nating series +1 −1 +1 −1 + . . . does NOT CONVERGE.
alterna
alternating
Depending on how we do the summation we have 3 possible answers : −1, 0, and
+1. So one can imagine the difficulties involved in a continuous summation with
continuous
UNCOUNTABLY many terms of both signs. Proving CONVERGENCE becomes very,
very difficult.
A more formal and rigorous treatment of INTEGRABLE dealing with the problems
of CONVERGENCE  discontinuities, finiteness and associativity  is covered
discontinuities finiteness
associativity
in A LITTILE MORE CALCULUS.
177 34. pplications
Integ
A pplica tions of Inte g r a tion There are numerous applications of in...
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 Fall '09
 TAMERDOğAN
 Limit, Δx

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