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I'm trying to find the double integral of dxdy over the area bounded by $y=\ln(x)$, $y=e+1-x$, and $y=0$.

I've drawn it out and I tried making the limits of $x: \ln(x) \to e+1-x$ and those of $y: 0 \to 1$ but I'm not getting the correct answer.

Vlad
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Zoe Pesce
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  • Hint, the lines intersect when $\ln(x)=e+1-x$. This has a unique solution $x=e$. As well, $\ln(x)=0$ when $x=1$ and $\ln(x)<0$ for $x<1$. Thus up to $x=1$, you only care about $y=e+1-x$ and $y=0$. When $1<x<e$, the area is bounded between $y=e+1-x$ and $y=\ln(y)$. – Alex R. Jul 12 '15 at 22:48

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Notice that for the region of integration, $0\le y\le 1$ and $e^y\le x\le 1+e-y$, so you can find

$\;\;\;\displaystyle\int_0^1\int_{e^y}^{1+e-y}1\; dx\;dy$.

user84413
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