Area between two curves with hyperbola - HELP please (1 Viewer)

[Lucas_]

Hey,

So the question is:

I keep getting my answer as Ln(2)- 4 units². Answer says Ln(2) + 0.5 units^2.


I don't get what I am doing wrong. Isn't the area between curves the integral between 0 and 2 of top curve - bottem curve? How do you even know which curve is top or bottem?

Thanks as always

 

[bashby]

draw the two graphs first, as y=1/x never reaches the y axis you cannot simply integrate from x = 0 to x=2. therefore to equate this you need to divide the area into a triangle (between y=x and the point where the two functions meet) and integrate the remainder of the area, therefore to find the area you need to integrate y=1/x between 1 and 2 and then add 1/2x1x1 (the area of the triangle)

 

[Lucas_]

I don't fully understand bashby, please explain?

I did draw them, but the line y=x means that I can integrate from 0 doesn't it?

Here, look:

 

[Carrotsticks]

 

[deswa1]

This isn't an area between two curves question. Consider the two areas (to the left of the point of intersection and to the right).

-x is less than 1: The hyperbola has no influence on this. You are trying to find the area of the triangle bounded by the origin, x=1 and y=1 (intersection of line and hyperbola). This is 0.5
- x is between 1 and 2: Here the area of the straight line is of no effect. You are trying to find the area of the curve y=1/x between 1 and 2 which is ln2-ln1=ln2

Add the areas together and you get ln2 + 0.5

 

[bashby]

If you were simply finding the area between the curve y=x and the x-axis then you could integrate from 0, but y=1/x never reaches 0 therefore you cannot integrate from 0 as you are finding an endless area, if you carry on with the hyperbole you will find it comes closer to the x-axis but never touches it. So you have to work out the area differently, with questions like this it is best to separate the area you are finding into two shapes, working one out without integration.

in this case the two lines meet making a triangle between the line y=x, the point of intersection and the x axis. therefore find where they meet (let 1/x = x then solve) then work out the area of that triangle adding it to the integration of y=1/x between 2 and the point of intersection.

 

[animerocker]

substitute y=x and y=1/x therefore you have to integrate from 1 to 2 as it's in the first quadrant. Another reason is that 0 is an asymptote therefore you cannot integrate from 0 to 2.
The upper curve is usually the one with a positive gradient and the bottom curve is normally a negative gradient.

 

[Lucas_]

Thanks to both bashby and Carrotsticks.

I now understand.

Can somebody please neg that Barbernator fellow?

 

[horny unicorn]

currrrrrves