Improper integrals and canceling of areas

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SUMMARY

The discussion focuses on the evaluation of improper integrals, specifically addressing the scenario of encountering the form "infinity minus infinity." It is established that this form does not equate to zero and indicates a divergent integral. The key takeaway is that when a function f has a singularity at a point b within the interval (a, c), the integral must be evaluated using limits independently, as shown by the expression \(\int_a^c f(x)dx= \lim_{\alpha\to b}\int_a^\alpha f(x)dx+ \lim_{\beta\to b}\int_\beta^c f(x)dx.

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When evaluating an improper integral and i get infinity minus infitiy when taking the limit. in what case do the areas cancel?
 
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Infinity minus infinity is not equal to zero. You have a divergent integral.
 
In general, no. If f has, for example, a singularity at x= b, where a< b< c, then [tex]\int_a^c f(x)dx= \lim_{\alpha\to b}\int_a^\alpha f(x)dx+ \lim_{\beta\to b}\int_\beta^c f(x)dx[/tex].

Those two limits have to be taken independently so you cannot cancel them.
 

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