How do you find the area between f(x)=10/x, x=0, y=2, y=10?

Jan 7, 2017

Area $= 10 \ln 5$

Explanation:

First examine the area drawn on a graph:

The bounded area, $A$ is made up of two components, the first is a rectangle bounded by $y = 2$, $y = 10$, $x = 0$ and $x = 1$, so it's area is given by:

${A}_{1} = 1 \times 8 = 8$

And the remaining Area, is that under the curve $y = \frac{10}{x}$, above $y = 2$ between $x = 1$ and $x = 5$. I found these values graphically, but algebra will also provide the values:

$\frac{10}{x} = 2 \implies x = 5$, and $\frac{10}{x} = 10 \implies x = 1$

This area is then given by:

${A}_{2} = {\int}_{1}^{5} \left(\frac{10}{x} - 2\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus = {\left[10 \ln x - 2 x \textcolor{w h i t e}{\int}\right]}_{1}^{5}$

$\setminus \setminus \setminus \setminus \setminus = \left(10 \ln 5 - 10\right) - \left(10 \ln 1 - 2\right)$
$\setminus \setminus \setminus \setminus \setminus = 10 \ln 5 - 10 - 0 + 2$
$\setminus \setminus \setminus \setminus \setminus = 10 \ln 5 - 8$

And so the total bounded area is:

$A = {A}_{1} + {A}_{2}$
$\setminus \setminus \setminus = 8 + 10 \ln 5 - 8$
$\setminus \setminus \setminus = 10 \ln 5$

Jan 7, 2017

$10 \ln 5 = 16.1$ areal units, nearly

Explanation:

graph{x(y-2)(y-10)(xy-10)=0 [-20, 20, -10, 10]}

The curved boundary is the branch of the rectangular hyperbola

xy = 10, in ${Q}_{1}$. See the graph.

The area is

$\int x \mathrm{dy}$, with x =10/y and y from 0 to 10.

$= 10 \int \frac{1}{y} \mathrm{dy}$, y from 2 to 10

$= 10 \left[\ln y\right] ,$ between y = 2 and y = 10

$= 10 \left[\ln 10 - \ln 2\right]$

$= 10 \ln \left(\frac{10}{2}\right)$

$= 10 \ln 5$ areal units.