<s>[INST] Finding the area between two equations The question asks to find the area of the part of the first Quadrant that lies between 
$x+4y-5=0$ and $xy=1$.
I began by first changing the equations to functions of $y$:
$x+4y-5=0$ to $y=\frac{5-x}{4}$ and $xy=1$ to $y=\frac{1}{x}$
I am not sure how to find the area, can you help me?
 [/INST] Set them equal to one another to find the intersection:
\begin{equation}
\frac{\left(5-x\right)}{4}=\frac{1}{x}\implies 5x-x^2=4\implies 0=x^2-5x+4,
\end{equation}
\begin{equation}
0=\left(x-4\right)\left(x-1\right)\implies x=\left\{4,1\right\}.
\end{equation}
And this matches the below graph:

Now  to find the area between them you must subtract the area under the lower curve from the area under the upper-curve:
\begin{equation}
\int\limits_{1}^{4}\left(\frac{5-x}{4}-\frac{1}{x}\right)dx=\int\limits_1^4\left(\frac{5}{4}-\frac{x}{4}\right)dx-\int\limits_1^4\left(\frac{1}{x}\right)dx=\left[\frac{5x}{4}-\frac{x^2}{2}-\log\left|x\right|\right]_1^4
\end{equation}
and I'll let you evaluate...

EDIT: 
You may also integrate with respect to $y$ by putting the values of $x$ back into the equations above and solving for the associated $y$ values and using those for your limits of integration. Then you will notice that the order of subtraction swaps, so you are finding the integral of the yellow curve in the graph above minus the integral of the blue curve. These will yield the same answer.
</s>