# Area between $y = \frac{1}{x}$, $y=\frac{1}{x^2}$, and $x = 2$

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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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? – Jordan Apr 28 '12 at 18:18

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 $x=2$.
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 $$\int_{x=1}^2 \left(\frac{1}{x}-\frac{1}{x^2}\right)dx.$$
@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)$which is wrong. – Jordan Apr 28 '12 at 19:09 @Jordan: I also get$(\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