Quote:
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Originally Posted by Dejhan_Tulip
What do you mean with that ?? Explain please...
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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.