Question

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

Answer 1

Area `= 10ln5`

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 xx 8 = 8`

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

`10/x=2 => x=5`, and `10/x=10=>x=1`

This area is then given by:

`A_2 = int_1^5 (10/x-2) \ dx`
`\ \ \ \ \= [10lnx-2x color(white)int]_1^5`

`\ \ \ \ \= (10ln5-10) - (10ln1-2)`
`\ \ \ \ \= 10ln5-10 - 0 + 2`
`\ \ \ \ \= 10ln5-8`

And so the total bounded area is:

`A= A_1+ A_2`
`\ \ \ = 8+10ln5-8`
`\ \ \ = 10ln5`

Answer 2

`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 dy`, with x =10/y and y from 0 to 10.

`=10 int 1/y dy`, y from 2 to 10

`=10[ln y],` between y = 2 and y = 10

`=10[ln10-ln2]`

`=10ln(10/2)`

`=10 ln 5` areal units.