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endeavor
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I'm learning about antiderivatives now, and I just learned a formula to find the antiderivative of xn:
if f(x) = xn, and F'(x) = f(x), then
F(x) = (xn+1)/(n+1) + C, where C is a constant
but if n < 0, then
F(x) = (xn+1)/(n+1) + C1, if x > 0
F(x) = (xn+1)/(n+1) + C2, if x < 0
My question is why are there two different C's?
The only example my book gives is of finding the antiderivative of f(x) = x-3. Its answer is:
F(x) = -1/(2x2) + C1, if x > 0
F(x) = -1/(2x2) + C2, if x < 0
EDITED: f(x) should be x-3
if f(x) = xn, and F'(x) = f(x), then
F(x) = (xn+1)/(n+1) + C, where C is a constant
but if n < 0, then
F(x) = (xn+1)/(n+1) + C1, if x > 0
F(x) = (xn+1)/(n+1) + C2, if x < 0
My question is why are there two different C's?
The only example my book gives is of finding the antiderivative of f(x) = x-3. Its answer is:
F(x) = -1/(2x2) + C1, if x > 0
F(x) = -1/(2x2) + C2, if x < 0
EDITED: f(x) should be x-3
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