Improper Integrals: Example 1: Divergent Sum

Math Easy Solutions

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In this video I go over an example on the first type of Improper Integrals, infinite intervals, and determine if the integral of the function 1/x from 1 to ∞ is divergent or convergent. I go over the solution to this and show that this is in fact divergent because the limit of the integral for the infinite interval does not exist as a finite number, but approaches ∞. In contrast to this, the previous function that I covered in my earlier video showed that the integral of the function 1/x^2 from x = 1 to ∞ approaches a finite number and converges to the value 1. This difference arises from the fact the function 1/x^2 converges to y = 0 much faster than for the curve 1/x as x approaches infinite. So in fact for the case of 1/x^2, as x approaches ∞ what really is happening is simply really small values are being added which only add to the number of decimal places and will continue to do so while approaching the value 1. On the other hand, the curve 1/x does not approach 0 fast enough for the sum to have a finite sum but rather keeps increasing the area for an infinite amount.

Download the notes in my video: http://1drv.ms/1Tmqrxz

View Video Notes on Steemit: https://steemit.com/mathematics/@mes/improper-integrals-example-1-divergent-sum