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Find the area bounded by $y=\operatorname{ln}(x)$ and $y=\operatorname{ln}^{2}(x)$?

Also how to find which curve will be above the other mathematically?

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  • $\begingroup$ How have you possibly got to integrating exponentials like this without going through derivatives of them beforehand? $\endgroup$ – Nij Jul 26 '17 at 10:46
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Find the intersection points: $$\ln{x}=\ln^2{x} \Rightarrow x_1=1,x_2=e.$$

Note: $$\ln^2x<\ln x, 1<x<e.$$

Hence: $$A=\int_1^e (\ln x- \ln^2 x)dx=\int_1^e\ln x dx -\int_1^e \ln^2 x dx=B-C$$ $$B=\int_1^e \ln xdx=x\ln x \bigg|_1^e-\int_1^e x\cdot \frac{1}{x}dx=1.$$ $$C=\int_1^e \ln^2 xdx=x\ln^2x \bigg|_1^e-\int_1^e x\cdot \frac{2\ln x}{x}dx=e-2B=e-2.$$ $$A=B-C=1-(e-2)=3-e.$$

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We have $ \ln x = (\ln x)^2 \iff x \in \{1,e\}$.

The area in question is given by $|\int_{1}^e (\ln x - (\ln x)^2) dx|$

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  • $\begingroup$ Why the absolute value around the integral? $\endgroup$ – Did Jul 26 '17 at 13:24

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