# Ask Uncle Colin: A fiddly Area

Dear Uncle Colin,

How would I find the area between the curve $y = \frac{1}{x^2}$, the coordinate axes, and the lines $y=4$ and $x=2$?

Don’t Really Appreciate What It Needs Graphically

Hi, DRAWING, and thanks for your message!

The sketch is the important thing here. We’ve got a shape with five boundary pieces: the $x$-axis from 0 to 2, the line $x=2$ from 0 to where it meets the curve at $\left(2, \frac{1}{4}\right)$, the curve from that point to where it meets the upper line at $\left(\frac{1}{2},4\right)$, the line from $\left(\frac{1}{2},4\right)$ to $(0,4)$ and the $y$-axis from $(0,4)$ to $(0,0)$.

The area of that regionsplits nicely into two sections: a rectangle with a base of $\frac{1}{2}$ and height 4 (so area 2), and an area under the curve we’ll need to integrate.

Using the limits we worked out (and writing $\frac{1}{x^2}$ as $x^{-2}$), that’s going to be $\int_{1/2}^{2} x^{-2} \dx$.

We get $\left[ -x^{-1}\right]_{1/2}^{2}$, which is $-\frac{1}{2}+2$, or $\frac{3}{2}$.

Overall, the area is $2 + \frac{3}{2} = \frac{7}{2}$.

Hope that helps!

- Uncle Colin