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I was given the problem: find the area of the region bounded by $y=1/x$, $y=x^2$, $y=0$, and $x=e$.

My approach was to break it up into two integrals, $\displaystyle \int_0^1 (x^2-0)\,dx$ and $\displaystyle \int_1^e\left(\frac{1}{x}-0\right)\,dx$.

Doing it this way resulted in an answer of 4/3. Simply by eyeballing it, the result seems reasonable, but I was wondering if there was a way to double check, or if someone else could verify that I am either doing this correctly or incorrectly.

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  • $\begingroup$ Thank you for cleaning my post up Arturo! $\endgroup$
    – Ryan
    Commented Feb 28, 2011 at 4:43

1 Answer 1

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What you did is indeed correct.

Splitting up as above, the area is $$A=\int_0^1x^2dx+\int_1^e \frac{1}{x}dx=\frac{1}{3}-0+\ln(e)-\ln(1)=\frac{4}{3}.$$

Hope that helps,

Eyeballing: So here is a way to see get be a bit more confident with the answer. When I drew the picture, it fit inside a rectangle of height $1$ and length $e$. (Vertices $(0,0)$, $(0,1)$, $(e,0)$, $(e,1)$). It looked like our area was roughly half of this rectangle, so $\frac{e}{2}$ since the rectangle has area $e\times 1=e$. On a calculator, $4/3=1.333$ and $e/2=1.359$ so we are pretty close, as expected.

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  • $\begingroup$ Thank you for the verification and the eyeballing technique! $\endgroup$
    – Ryan
    Commented Feb 28, 2011 at 3:48
  • $\begingroup$ @Ross: For some reason whenever I type "e" inside dollar signs, that is $e$, it becomes an epsilon. $\endgroup$ Commented Feb 28, 2011 at 3:48

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