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I have no idea how to do a problem like this, the answer seems to be infinity to me. I am asked to find the area between the curves:

$y = \frac{1}{x}$

$y=\frac{1}{x^2}$

$x = 2$

I have no idea what this means. I graphed it and it didn't help. If the graph stops at $2$ that doesn't really help me out at all. It looks like I have infinity as the answer.

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    But what about negative numbers?2012-04-28

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The problem (in Stewart) asks you to find the area of the region enclosed by the curves $y=1/x$, $y=1/x^2$, and $x=2$.

First draw a careful diagram. There are at least two regions that could be thought of as enclosed by the curves. However, there is only one finite region. At that stage of the book, areas of infinite regions (improper integrals) have not been introduced. So the natural interpretation is that the finite region is intended.

The first two curves meet at $(1,1)$, and for $x>1$, the curve $y=1/x^2$ lies below $y=1/x$. After you have identified the region that we want to find the area of, it should not be hard to see that this area is $\int_{x=1}^2 \left(\frac{1}{x}-\frac{1}{x^2}\right)dx.$

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    @Jordan: I also get $(\ln 2 +1/2)-(0+1)$. This simplifies to $\ln 2 -1/2$, which is the answer given in the back of the book, at least in my edition. Note that $2^{-1}-1=\frac{1}{2}-1=-\frac{1}{2}$.2012-04-28