# Area between two curves

• ashleyk

#### 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...

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.

The attached plot is the area. It is:

$$\iint_R dA$$

Right?

#### Attachments

• area5.JPG
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and what about area between x=0 and x=1?

cronxeh said:
and what about area between x=0 and x=1?

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

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.

saltydog said:
The attached plot is the area. It is:

$$\iint_R dA$$

Right?

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

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

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

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..

i don't 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 can't really find an area 'between' curves per se- as there are interlapping curves

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

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).

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.