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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:

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


$$x = 2$$

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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The curves intersect at x=1. – Matthew Conroy Apr 28 '12 at 18:05
The three curves $y=1/x,y=1/x^2$, and $x=2$ bound a more or less triangular region with one straight and two somewhat curved sides. The straight side, of course, is $x=2$ between $y=1/2$ and $y=1/2^2=1/4$. Where do the other two sides meet? (See also @Matthew’s comment.) – Brian M. Scott Apr 28 '12 at 18:11
But what about negative numbers? – user138246 Apr 28 '12 at 18:18
up vote 6 down vote accepted

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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But in the negative region the graph gets pretty ridiculous, what do I do? – user138246 Apr 28 '12 at 18:19
@Jordan: It’s always understood in these problems that you use the finite region bounded by the given curves. In this case that region lies entirely between $x=1$ and $x=2$. – Brian M. Scott Apr 28 '12 at 18:22
I still can't get the answer I am left with (ln2 + 2^{-1}) - (0 + 1)$ which is wrong. – user138246 Apr 28 '12 at 19:09
@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 $2^{-1}-1=\frac{1}{2}-1=-\frac{1}{2}$. – André Nicolas Apr 28 '12 at 19:18

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