0

Find its value

$$f(x)=\frac{1}{x^2}, I = [-1,1]$$

obviously this is undefined at $$x=0$$

$$\int_{-1}^0 \frac{1}{x^2} + \int_0^1 \frac{1}{x^2}$$

set $0 = b$

for the first integral:

$$\int_{-1}^b \frac{1}{x^2} = \lim_{b\to0^-} \frac{-1}{x} \Big\vert_{b}^{-1} =$$ undefined $-(-1)$

for the second integral: -1-undefined. The back of the book says its just undefined

$$\int_0^1 \frac{1}{x^2}=\lim_{b\to 0^+} \frac{-1}{x} \Big\vert_b^1$$

K. Gibson
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  • See: https://math.stackexchange.com/a/2098557/693161 – Andrew Chin Nov 22 '19 at 17:25
  • thats way too long of a proof for this class its an introduction to real analysis, in the past when I have used longer and more complicated proofs from answers I get marked off significantly – K. Gibson Nov 22 '19 at 17:29
  • Perhaps if that was mentioned, we would've just known to say "the integral diverges". – Andrew Chin Nov 22 '19 at 17:31
  • In your evaluation of the first integral, you've incorrectly switched the lower and upper limit of integration.
  • Once you've shown one of the two diverges, it follows that the entire integral diverges. No need to check the other integral.
    • – bjorn93 Nov 22 '19 at 20:52
  • Is your question why is it when you find "undefined $\ -(-1)$" and "$-1 -\ $ undefined", the book just says "undefined"? That is because "undefined" is not a value. You cannot add to it or subtract from it. It just means that there is no number the limit converges to. "undefined $\ -(-1)$" and "$-1 -\ $ undefined" are expressions without any meaning. That is, they are "undefined". – Paul Sinclair Nov 23 '19 at 01:27