Integrate Signed Attempt with airSlate SignNow

Integrate signed attempt

good afternoon I'm Mr ish thank you for joining me for this video we have a good verification procedure here for us to do and let's get right into it we have in this sentence if a comma B are real numbers and we're seeing the a right here and the B right here as part of our interval limits if a and b are real numbers and this integral from minus infinity to positive Infinity f of x DX is convergent this convergent here is a more important word we have to show this is equal to that from minus infinity to a f of x DX plus a to Infinity f of x DX is equal to minus infinity to be f of x DX plus b Infinity f of x DX when I'm seeing something right here in terms of my integral form and I'm seeing the word here convergence I'm really thinking about type 1 improper integral there's nothing here mentioned in with regards to a infinite discontinuity nothing here mentioned with regards to an AS him toward a vertical asymptote we're looking here at a clear cut case of a type 1 improper integral very likely the fact that we're seeing this type of presentation here and a good convergence we're probably looking at something in terms of a graph it doesn't have to be an accurate depiction but it could very well be something like this going here towards infinity and minus infinity all of this area right here is very well being evaluated by this expression here from minus infinity to Infinity it's convergent which means an area of determination is possible and limits exist at this end and limits exist here limit as X approaches minus infinity f of x approaches very likely this horizontal asymptote limit as X approaches Infinity the f of x here approaches this horizontal asymptote which is your x-axis y equals zero so in all instances here we have a clear-cut case of convergence and the fact that we have the placement of these A and B what does that really mean it doesn't really mean anything too significant a for all purposes could have very well been zero you could be doing minus infinity to a and zero and then zero to Infinity B could be any minus number it could be any positive number then you're doing minus infinity to B and then B to positive Infinity the point here is that placement of A and B it doesn't affect anything here with regards to how you place these specific limits because each of these limits are shared across their intervals and everything works out in terms of convergence anyhow let's get back to this question and let's prove this to be the case and it's not hard you can prove this very quickly by applying limits you can do over here the simple type 1 and proper integral procedure limit as T approaches is...