# Area between two curves

1. Apr 17, 2005

### ashleyk

Find the area of the region bounded by the curves

y=x , y=1/(x^2) , and x=2

I know after you sketch it you have to take the intergral of the top function minus the bottom function from the points that they intersect. I am stuck however because one function does not appear to be on top throughout the interval. Any help would be great...

2. Apr 17, 2005

### Moo Of Doom

I'm thinking that since there was no other bound (you can't assume the y axis is a bound), I imagine their point of intersection is meant to be taken as the other bound for the function.

3. Apr 17, 2005

### saltydog

The attached plot is the area. It is:

$$\iint_R dA$$

Right?

#### Attached Files:

• ###### area5.JPG
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4. Apr 17, 2005

### cronxeh

and what about area between x=0 and x=1?

5. Apr 17, 2005

### saltydog

Ok, I see what you mean. Perhaps the problem is not well defined.

6. Apr 17, 2005

### HallsofIvy

Staff Emeritus
The region is bounded by y= x, y= 1/x2, and x= 2.
The only difficulty with that is that there is no lower boundary. I bet the problem also gives y= 0 as a boundary. In that case:

Of course, y= x and y= 1/x2 cross at (1, 1). That is, the region is bounded above by y= x from x=0 to x= 1, then by 1/x2 from x= 1 to x= 2.

The simplest way to do this is treat it as two separate regions:
1) the region bounded by y= x, y= 0, x= 1. In fact, that's a triangle with area 1/2.

2) the region bounded by x= 1, y= 1/x2, y= 0, x= 2.
Integrate y= 1/x2= x-2 from x= 1 to x= 2.

Add the areas of the two regions.

7. Apr 17, 2005

### Jameson

I think this is a Calc I question, so the poster wouldn't have encountered double integrals yet.

8. Apr 17, 2005

### ashleyk

I got an answer of 1? can anyone verify...??

9. Apr 17, 2005

### HallsofIvy

Staff Emeritus
How did you get 1? In fact, can you tell me what the bottom curve of your region is?

10. Apr 17, 2005

### ashleyk

I got the answer of one assuming the bottom is y=0...i have to ask my teacher tomorrow for sure if that is suppose to be one of the curves..

11. Apr 17, 2005

### cronxeh

i dont know how many curves you people see, but there were (from top of my head - i plotted this in the morning) 4 different regions. you have one between x=0 and x=1 - the one where 1/x^2 reaches to infinity, area is approximately 144+, then between x=1 and x=2 there is a triangular area and a parabolic area over on top.

so this question is not very well defined. you cant really find an area 'between' curves per se- as there are interlapping curves

12. Apr 18, 2005

### Moo Of Doom

I'm 99% sure the region is meant to be this one (in really crappy, fast sketching)...

#### Attached Files:

• ###### graph.jpg
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13. Apr 18, 2005

### IntuitioN

the intersection of y=1/(x^2) and y=x is at x=1

so the area is the area under the top curve - area under the bottom curve.

A = [I(1 to 2)(x dx)] - [I(1 to 2)((dx)/(x^2))]

I = integral sign

14. Apr 18, 2005

### HallsofIvy

Staff Emeritus
Yes, or course! to both Moo of Doom and IntuitioN. I didn't draw my graph clear enough!

You could also just integrate the difference of the y-values
$$A= \int_1^2 (x- x^{-2})dx$$
(that's exactly the same as what IntuitionN gave).

15. Apr 22, 2005

### abia ubong

hey,
its obvious from a book i read u equate x and 1/x^2,getting x as 1.this ends up being the lower boundary number,and then integrate definitely the multiple of x and 1/x^2gettiung the integral of 1/x,which becomes log [x].taking the definite step, the answer is log2,in base e.