ecoli Posted March 26, 2005 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.
mezarashi Posted March 26, 2005 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.
ecoli Posted March 27, 2005 Author 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?
Dave Posted March 27, 2005 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.
Dave Posted March 27, 2005 Posted March 27, 2005 Not a problem. Hopefully back up today, if not then tomorrow.
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