I get a $0$ for the area when I know it's supposed to be a positive value: $$\int_0^e \ln(x) \,dx $$ $$\int \ln(x)\,dx = x\ln(x) - x = [e\ln(e) - e] - [0\ln(0) - 0] = [e - e] - [0] = 0$$ How did I go wrong? I know the actual area can't be equal to $0$.
Find the $\int_0^e \ln(x)\,dx =0$?
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calculus
definite-integrals
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0https://www.wolframalpha.com/input/?i=integral+ln(x)+from+0+to+e "I know the actual area can't be equal to 0." Why not? – 2017-02-07
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0You have it correct. – 2017-02-07
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0It's signed area, area below $x$ axis counts as negative. – 2017-02-07
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0If you wanted area, then you need to integrate $\lvert \ln x\rvert $, not $\ln x$. – 2017-02-07
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0@kobe Or, equivalently, $\int_1^e \ln x\,dx - \int_0^1 \ln x\,dx$. – 2017-02-07
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2 Answers
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Why couldn't it be equal to $0$? You have $\ln x < 0$ for $x\in(0,1)$, and $\ln x > 0$ for $x\in(1,e]$; thus, $$ \int_0^e \ln x \,dx = \underbrace{\int_0^1 \ln x \,dx}_{< 0} + \underbrace{\int_1^e \ln x \,dx}_{> 0} $$ and the fact that the two terms cancel is not a contradiction to anything.
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0@Michael Hardy: thanks for fixing my $\LaTeX$ blunders :) – 2017-02-07
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Your answer is correct. If you think about the log function, it takes on negative values for all arguments less than 1, so it kind of makes sense. The main error you made, however, was taking the $\ln(0)$ which is negative infinity. What you need to think about is taking a limit as $x\to0$, see if that helps.