alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

Fx element say fx x and then see that indeed fx dx

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Unformatted text preview: t is easy to see that the energy E, as a function of displacement, will energy be infinite. And hence: E = ∫ m.a.ds will not CONVERGE. m.a.ds Here we implicitly assumed that the length of the pendulum was infinite. What if the pendulum was of finite length ? Does it mean that the pendulum will swing all the way round in a circular orbit ? What if the ball bounces indefinitely over infinite time interval [0, +∞) with gain in energy on each bounce ? Will the ball disappear both in TIME and SPACE ? It is of fundamental importance to see the relationship between the abstract calculation on the one hand and the physical meaning ( the reality - what is happening in nature) on the other hand. With all this mind we now give informal definition of INTEGRABLE. 1. f(x) is defined over [a, b] (that is to say SINGLE VALUED) except at the points of discontinuity. And f(x) has at most COUNTABLY many discontinuities over [a, b]. So between each pair of consecutive such points of discontinuity we have a sub-interval. 2. An each such point of discontinuity f(x) has 2 definite values. discontinuity 3. continuous summation Over each of the sub-intervals in [a, b] the continuous summation contin must be something FINITE, say : S0 , S1 , S2 , . . . , Sn , . . . . 4. The combined discrete summation S = { S0 + S1 + S2 + . . . + Sn + . . . } discrete ∫f(x)dx over all the COUNTABLY many sub-intervals must also be something FINITE b so that ∫ f (x)dx CONVERGES. a Then we say f(x) is INTEGRABLE. 174 A word on CONVERGES The fundamental requirement in integration is that the continuous summation integration continuous CONVERGES. Let us compare two examples of discrete summation and continuous discrete continuous summation . The discrete summation discrete +1 ∞ 1 Σ /n does NOT CONVERGE. n=0 The integral ∫ { 1/x} dx is a continuous summation over the finite interval [0, +1]. continuous summation contin 0 But the summation does not CONVERGE. So it is not INTEGRABLE. +∞ f(x) = 1 x / 2 1 ½ 0 −∞ ½1 2 ... +∞ −1 −2 −∞ We can see from the diagram that the area under the curve f(x) = 1 x over the finite / inter...
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