Thread: .999~ = 1
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Unread 03-03-2004, 17:34
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Re: .999~ = 1

Quote:
Originally Posted by Dejhan_Tulip
What do you mean with that ?? Explain please...
essentially, an indefinite integral will always be equal to the same indefinite integral plus a constant. it would have to be, as the following example shows:

f(x) = 2x + 1
g(x) = 2x

f'(x) = 2
g'(x) = 2

if we didn't have a constant of integration,
integral(f'(x)) = 2x = integral(g'(x)) = 2x
and thus f(x) would be implied to be = to g(x)

however, the correct form is this:
integral(f'(x)) = 2x + C1
integral(g'(x)) = 2x + C2

thus, f(x) = g(x) + C2 - C1, where C2 and C1 are integers.

essentially what this all means is that you can't simply "take out" an indefinite integral from an equation, and thus the fact that 0 + integral(f(x)) = 1 + integral(f(x)) does not imply that 0 = 1.
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