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cabellos6
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Homework Statement
the integral of 3/x
Homework Equations
The Attempt at a Solution
am i right in saying this is 3lnx
Is the Integral of 3/x Equal to 3lnx?
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SUMMARY
The integral of 3/x is definitively equal to 3ln|x| + C, where C represents the arbitrary constant. When evaluating the improper definite integral from 0 to infinity, it requires limits at both endpoints, specifically 3∫_0^{∞} (1/x) dx, which can be expressed as 3lim_{a → 0+}∫_a^{1} (1/x) dx + 3lim_{b → ∞}∫_1^{b} (1/x) dx. The discussion emphasizes the importance of absolute value bars in the logarithmic expression and the necessity of proper limit handling for improper integrals.
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the integral of 3/x
am i right in saying this is 3lnx
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Kurdt
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Yes you are. Don't forget the arbitrary constant.
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quantumdude
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And don't forget the absolute value bars. It's 3ln|x|+C.
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Kurdt
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Tom Mattson said:
Indeed
Welcome back Tom!
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rxtrejo
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Same question adding Limits of integration b=infinity a=0
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Hootenanny
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Well, what do you think it is?rxtrejo said:
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NJunJie
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rxtrejo - you asking a question?
That would mean substituting the bounded values and find some 'area' within the limits you have given.
That would mean substituting the bounded values and find some 'area' within the limits you have given.
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Mark44
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Not so fast. This is an improper definite integral that requires limits at both endpoints to evaluate.NJunJie said:
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Mark44
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IOW,
3\int_0^{\infty} \frac{dx}{x}<br /> = 3\lim_{a \rightarrow 0_+}\int_a^{1} \frac{dx}{x}~+~ 3\lim_{b \rightarrow \infty}\int_1^{b} \frac{dx}{x}
I chose to split the first integral at 1. Any reasonable value could be used to divide the original interval into two subintervals.
3\int_0^{\infty} \frac{dx}{x}<br /> = 3\lim_{a \rightarrow 0_+}\int_a^{1} \frac{dx}{x}~+~ 3\lim_{b \rightarrow \infty}\int_1^{b} \frac{dx}{x}
I chose to split the first integral at 1. Any reasonable value could be used to divide the original interval into two subintervals.
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