ecoli Posted March 26, 2005 Share Posted March 26, 2005 [math] $If$ [/math] [math] \int_a^b f(x) dx = a +2b [/math] [math] $then$ [/math] [math]\int_a^b (f(x) +5) dx = ?[/math] How do you work this one out? The "+ 5" part is really screwing me up. Link to comment Share on other sites More sharing options...
mezarashi Posted March 26, 2005 Share Posted March 26, 2005 After a year or so of not having done calculus I hope my fundamentals still hold. If you consider the property of linearity of the integration, then you can always simplify the second equation to: int(b,a) [ f(x) + 5]dx ---> int(b,a) [f(x)]dx + int(b,a) [5]dx I guess the answer from here is obvious. The first term leads to (a + 2b) and the second term leads to 5x, substituting in b and a to get: 5b - 5a. Link to comment Share on other sites More sharing options...
ecoli Posted March 27, 2005 Author Share Posted March 27, 2005 Oh no...can somebody check this. I posted this question on another math forum and this is the answer they gave. [math] \int_a^b(f(x)+5)dx=\int_a^bf(x)dx+\int_a^b5dx=a+2b+[5x]_a^b=a+2b+5b-5a=7b-4a [/math] Which one's right? edit: - why is Latex not working? Link to comment Share on other sites More sharing options...
Dave Posted March 27, 2005 Share Posted March 27, 2005 They're both right: mezarashi was saying that first integral is equal to (a+2b) and the second is equal to (5b - 5a), so adding them together gives you the correct answer. It's just that mezarashi's post wasn't worded in the best way possible LaTeX is down because of a rather nasty bug that I found in the software. It's fixed, but I'm waiting for blike to get online so I can put the new version up. Link to comment Share on other sites More sharing options...
ecoli Posted March 27, 2005 Author Share Posted March 27, 2005 Thanks and Thanks, Dave! Link to comment Share on other sites More sharing options...
Dave Posted March 27, 2005 Share Posted March 27, 2005 Not a problem. Hopefully back up today, if not then tomorrow. Link to comment Share on other sites More sharing options...
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