I need a little help with a Calculus problem. Here is the original problem.

Here is my work thus far. I know that the third graph is correct already. I can't figure out why 20/7 is incorrect though. I keep getting the same answer no matter how I attack the problem.

 A: Your integrand should be $9 -1/x^2$.  That's the height of the Riemann rectangle.  when you just put $1/x^2$ you're getting the area under the blue region, rather than the blue region.
A: Goddard's method is correct. Another way to think about it is that the area inside the 7 by 9 rectangle is comprised of three sums:
$$63 = 3 + \int_{\frac{1}{3}}^7 \frac{1}{x^2}\mathrm{d}x + R$$
where $63$ is the area of the 7 by 9 rectangle, 3 is the area of the $\frac{1}{3}$ by $9$ strip in the left of the rectangle, the integral is the area under the curve, and $R$ is the area of the region you desire. You've already calculated the integral as $\frac{20}{7}$, so plugging in we find $R = 60 - 20/7 = 400 / 7$. 
A: If you consider the full rectangle enclosed by the two axes, $x=7$, and $y=9$, its area is $63$ units. Now, a reasonable guess of the area $R$ might be some 80% of the rectangle, which would then be roughly $50$. Now, $20/7$ is roughly $3$. The discrepancy here should be enough to show you why $20/7$ is intuitively incorrect.
Now, to calculate the area $R$, you must pick a valid integrand and the proper bounds. Your bounds seem to be correct, and we will integrate from $x=\frac{1}{3}$ to $x=7$. The upper function between these two $x$ values is $y_{\mathrm{\,upper}}=9$ and the lower function is $y_{\mathrm{\,lower}}=\frac{1}{x^{2}}$. So, the distance between $y_{\mathrm{\,lower}}$ and $y_{\mathrm{\,upper}}$ is
$$\begin{equation}\begin{split}
y & = y_{\mathrm{\,upper}} - y_{\mathrm{\,lower}}
\\& = 9 - \frac{1}{x^{2}}
\end{split}\end{equation}$$
and so our integral is
$$\begin{equation}\begin{split}
R & = \int_{\frac{1}{3}}^{7}\left( 9-\frac{1}{x^{2}} \right)dx \\
& = \left[ 9x+\frac{1}{x}\right]_{\frac{1}{3}}^{7} \\
& = \left( 63 + \frac{1}{7} \right) - \left( 3 + 3 \right) \\[4pt]
& = \frac{400}{7} \approx 57.14.
\end{split}\end{equation}$$
Roughly $57$ is a much more reasonable computation of $R$.
A: 
Notice, the area required is bounded by the curves $y_1=7$ & $y_2=\dfrac{1}{x^2}$ from $x_1=\dfrac13$ to $x_2=7$ (As shown in above figure)
Now, consider a vertical rectangular slab of differential area $(y_1-y_2)dx$ (as labeled by green color in above figure). The total area of bounded region is given by integrating the differential area $(y_1-y_2)dx$  with proper limits as follows 
$$\int (y_1-y_2)dx=\int_{1/3}^{7}\left(9-\frac{1}{x^2}\right)dx=\frac{400}{7}\ \mathrm{unit}^2$$
