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Area enclosed by two curves

  1. Jan 29, 2008 #1
    Find the area of the region enclosed by the curves, and decide whether to integrate with respect to x or y. y=3/x, y=6/x^2, x=5
    anyone able to explain how to approach a problem like this I've tried it a few times and get the wrong answers but i dont even know what kind of answer im supposed to get like just a number, or a function, something with variables, any help would be great thanks ahead of time.
  2. jcsd
  3. Jan 29, 2008 #2
    you are supposed to get a number. Try to draw the three lines in a (x,y)-coordinatesystem, then you see that the lines enclose a area. Then you integrate 1 over this area to get the area.

    A quick example:

    lets say you have the triangle given enclosed by the a-axis, the y-axis and the line y=1-x. This corespons to the triangle with coners (0,0),(0,1) and (0,1) (again if in doubt draw the line in a coordinatesystem). This triangle you know have area ½, because you know the area formula of a triangle but you can always get it by intagration.

    [tex] \int_0^1 \int_0^{y=1-x} 1 dy dx = \int_0^1 [y]_0^{y=1-x} dx = \int_0^1 1-x dx = [x-1/2x^2]_0^1 = 1-1/2 \cdot 1^2 = 1/2 [/tex]

    think of it as you choose a fixed x on the x-axis, then integrate along the y-axis, when you draw the triangle you clearly see that you have to start at 0 always, but depending where you are on the x-axis (the x you chose), you need only to go up to y=1-x or else you get out of the triangle, next when you integrate along the x-axis you "add" togheter the contribution of each of the lines from 0 to y=1-x, a contribution from each x you choose between 0 and 1 on the x-axis (hope this makes sence to you).

    As you see i decided to integrate with respect to y first because then i didn't have to isolate x in y=1-x, but i could just aswell have written x=1-y and then

    [tex] \int_0^1 \int_0^{x=1-y} 1 dx dy = \int_0^1 [x]_0^{x=1-y} dy= \int_0^1 1-y dy = [y-1/2y^2]_0^1 = 1-1/2 \cdot 1^2 = 1/2 [/tex]

    the reason you are asked to decide wether to integrate with respect to x or y first is that sometimes you can't peform the integral if you choose the wrong first, fx. it could be impossible to isolate x in the equation, or when you do it the one way around you simply get something that can't be integrated analytically. If you can't figure out what to integrate first, then try to do both, because it's a good exercise to see what goes wrong if you do the wrong thing.
  4. Jan 29, 2008 #3
    I dont know I cant seem to get the answer there's something im doing wrong i use the equations to find the bounds of integration which i think are x=2 to x=5 if not x=2 then x=0 i've tried with all 3 numbers though, anyway I then integrate the 2 equations y=3/x and y=6/x^2 (six over x squared) and from there i plug in two of the numbers and subtract the differences to get a number and can't get it right. i've tried every which way with 2 of the 3 x values i gave above and even tried adding instead of subtracting incase that isnt the way your supposed to do it, but I still haven't found the right answer. Can anyone see where my mistake is?
  5. Jan 30, 2008 #4
    i not surewhat you are doing, but it seem like you misunderstood what you are suppose to do. You should not integrate over the functions, the functions enclose a area in the xy-plane, and then the area of this area. As you can see you need you bounds of integration to variable, because you don't enclose a square.
  6. Jan 30, 2008 #5


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    Have you graphed the functions so you can see the area? The highest and lowest x (integrating with respect to x) and y values (integrating with respect to y) will tell you what your limits of integration should be. In particular, where do those graphs cross?

    And surely you know that the "area of a region" is a number!
  7. Jan 30, 2008 #6
    I have graphed the functions and the only time they cross seems to be at x=2 they have the same y value but then after that they don't cross again which makes me think where the x=5 comes into play
  8. Jan 31, 2008 #7


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    So the only "region" bounded by those graphs is between x= 2 and x= 5. (You need to graph all three, y= 3/x, y= 6/x2, and x= 5 to see the region.)

    Now, between x= 2 and x= 5, which curve is higher? Each of the "rectangles" approximating the area will have height (y2- y1) where y2 is the larger of the two values.
  9. Jan 31, 2008 #8
    okay, you are right that the integral in x, should run from 2 to 5. But lets look at how y should run for a given x in [2,5]. So fix x=3 (see attachment) then

    y = 3/x = 3/3 = 1

    y= 6/x^2 = 6/3^2 = 6/9 = 2/3

    so for x=3, y should run from 2/3 to 1. But for x = 4 we have

    y = 3/x = 3/4

    y= 6/x^2 = 6/4^2 = 6/16 = 3/8

    so for x=4, y should run from 3/8 to 3/4.

    or in generel for x, y should run from y= 6/x^2 to y= 3/x you see?

    lets see on the integral along y, from y= 6/x^2 to y= 3/x.

    [tex] \int^{y=\frac{3}{x}}_{y=\frac{6}{x^2}} dy [/tex]

    this is a function of x, so
    [tex] f(x) = \int^{y=\frac{3}{x}}_{y=\frac{6}{x^2}} 1 dy [/tex]
    so for every x you get the area under the constant function 1, from y=3/x to y=6/x^2. Try to look at the second picture in the attachment, you se that this is the area under one of the lines i drawn. So by "adding" the area of all these lines for every x, you get the volumes of the shaped triangle drawn in the bottom corner. But this is exactly the area of the "triangle" on the first picture because the shaped triangle exactly have the height 1.

    So now how do we "add" all the areas? By this we meen integrate over x so we get

    [tex] area = \int_2^5 f(x) dx = \int_2^5 \int^{y=\frac{3}{x}}_{y=\frac{6}{x^2}} dy dx = \int_2^5 \frac{3}{x}-\frac{6}{x^2} dx = [3Ln(x)+\frac{6}{x}]_2^5 = -\frac{9}{5}+ Ln(\frac{125}{8}) \approx 0.9489[/tex]
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