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Integral, Triangle areas?

  1. Oct 14, 2003 #1
    OK, say a 90deg triangle, with 5, 12, 13.

    We know area of it is 30.

    because 1/2bh... now how do u express it in integrals?

    And how do you find the value of y at which the triangle is divided into 2 equal sections?
  2. jcsd
  3. Oct 15, 2003 #2


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    find the area under the line

    y = 5x/12

    between x=0 and x=12

    [inte] 5x/12 dx = 5x2/24

    evaluate at the upper limit (lower limit yields zero)

    =5(144)/24= 30

    I am not quite sure what you mean by the second part, put suddenly I feel as if I have done your homework, so.... I'll let you puzzel out the second part, haveing seen how I did the first part.
  4. Oct 15, 2003 #3
    Learning is my homework :p

    what is 5x/12? bh/2, isn't that the area of the triangle?

    the 2nd part is what I wanted to know a long time for a real life problem. I have a garden, with a shape like a triangle. I was thinking of adding a wall to it so that it makes 2 sections, 1 for vegetables and 1 for others, but both have the same area.

    The problem I am thinking might relate to calculus, that's what I heard, ... But I don't know how to use calculus to find where should i start (0,y) to add the wall...
  5. Oct 15, 2003 #4
    ok, i see why you said find area under 5/12x...
  6. Oct 15, 2003 #5


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    If you look at the expression I derived,

    A = 5/24 x2

    You have an expression for the Area, given x, what you want to do is find x where the Area = 15

    15 = 5/24 x2
    x = sqrt (72)= 6 sqrt(2)

    This is the distance from the smallest angle along the long (not hypotenuse) side.

    If that not the side you want the wall from, simply rewrite the equation for the line, only with the approiate slope you orient your garden correctly.

    (Does that make sense?) :)
  7. Oct 15, 2003 #6
    ok, I get the concepts part,

    can you explain what the "dx" part mean in the integral?

    My understanding for integral is not very good, only know how to solve a few of them, but I would like to know what the dx parts mean so that I can make more use of it.

    And we took the integral of the line y = 5/12x,... that gives us the area. So that means the derivative of area is a line that borders the area??

    also, what is the "x" in the integrated equation?

  8. Oct 15, 2003 #7
    and how do you find at what "y" value(the height in this case), does it make the triangle 2 equal sections?(can't be 6sqrt(2), because the height is 5)
  9. Oct 15, 2003 #8


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    Yes, you really are weak on integrals. It might pay you to go back and review "Riemann sums".

    One way of defining "integral" is that ∫ab f(x) dx IS the area between x= a and x=b bounded on top by the graph of y= f(x) and below by the line y= 0.
    One way of approximating that is to divide the x axis into many pieces, each of length Δ x. Construct a rectangle on each piece with height f(x*) where x* is some x value inside the piece and that will be close to the area of the curve above the piece. The sum of the areas of all such rectangles will be close to the area under the entire curve. One can show that taking the limit as the number of pieces goes to infinity give the exact area. That is written as ∫f(x) dx where the ∫ represents the Σ of the sum and dx represents the Δ x.

    In your case, a 90 degree triangle with legs of length 5 and 12 DOESN'T HAVE any graph or equation connected with it so you CAN'T find the area by an integral!

    Well, you can, IF you place a coordinate system on it so that you can think of it as a graph. Staring at a right triangle for a very long time (I'm kinda slow!), it starts to dawn on me that the two legs look a little like x and y axes so I consider setting up a coordinate system with the right angle at (0,0), the x-axis along one leg, and the y-axis along the other. Now I notice that the hypotenuse is (part of) a line passing through the points (0,12) and (5,0) (if you set up your coordinate system so that it passes through (12,0) and (0,5), Okay, that'll work too.).

    The line passing through both (0,12) and (5,0) has slope (12-0)/(0-5)= -12/5 and so its equation is y= -(12/5)(x-0)+ 12 or y= -(12/5)x+12. (Integral set up his coordinate system so that leg with length 12 was the x-axis and the leg with length 5 was the y-axis. He got y= -(5/12)(x-12) as the equation of the line {though he mistakenly wrote it as y= -(5/12)x!!). Okay, that'll work too.)

    The area is, according to the theory, ∫05 (-(12/5)x+ 12)dx. The "antiderivative of this is -(6/5)x2+ 12x+ C. Evaluating at x=5 and x= 0 we get -(6/5)(25)+ 12(5)+ C= -30+ 60+ C= 30+ C and -(6/5)(0)+ 12(0)+ C= C. Subtracting:
    30+ C- C= 30. Exactly the same as (1/2)(5)(12)!!!!
  10. Oct 15, 2003 #9
    Yes, my integral is not good at all, that's why I post here to seek for help.

    I never understood the LRAM or whatever it is called approximation method, is that important?

    Never looked at Rieman sums either:\

    However many way you approached the triangle problem, I thought it can be modeled as integral of 5/12x dx because the area is between the line y=5/12x and y=0.

    Thus integral of 5/12x - 0 = integral of 5/12x dx.

    And would someone please answer my questions prior to his response pls.
  11. Oct 16, 2003 #10


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    I choose the simpler line. Place the origin, not at that right angle but at the either of the other angles so the hypotenuse passes through the origin. Now the hypotenuse IS the line we wish to find the area below. Since the line passes through the origin (by design) we simply use the length of the other 2 sides as defining the slope, using rise over run (y/x). The line is given by y=mx you now simply integrate the equation of the line with limits of the origin to the length of the side on the x axis.

    Thus the integral yields a slightly different, but equivalent, to the standard expression for the area of the triangle, it is different because it incorporates our knowledge of the relationship between the 2 sides.

    in general the area = mx2/2

    where m is the slope of the hypotenuse which must pass through the origin.
    There is no mistake! The line I used is y=5/12 x.

    Last edited: Oct 16, 2003
  12. Oct 16, 2003 #11


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    OPPs, missed this,
    The dx is simply notation, it must follow the integral symbol, view it as a form of a period, it blocks off the function to be integrated.

    The x in my expression

    A = mx2/2 (see previous post) is the distance from the origin along the leg that forms the x axis. So if you placed your wall perpendicular to the long leg 6Sqrt(2)~8.5m from the small angle you will split your garden in half.

    Edit: I missed the square in my exprssion for area.
    Last edited: Oct 16, 2003
  13. Oct 16, 2003 #12
    but the width(height) is only 5 meters total.
  14. Oct 16, 2003 #13


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    One http://home.comcast.net/~rossgr1/Math/Garden.pdf is worth a thousand words.

    I have a feeling you are going to say, "But that is not how I wanted the wall to run!"

    Look back at my first post I said
    You never bothered to clear that up, you only spec was to split the garden in half. The dotted line does that.
    Last edited by a moderator: Apr 20, 2017
  15. Oct 17, 2003 #14
    ok, can i plug that back into bh/12 =15, where b is 6sqrt(2) and find h?
  16. Oct 17, 2003 #15


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    Yes if you mean bh/2=15,Or you could plug it into 5x/12 where x=6 sqrt(2), or you could read it off my graph ~3.5
    Last edited: Oct 17, 2003
  17. Oct 17, 2003 #16
    What if I didn't use a very straight wall? Say I used a sine shaped curve as my wall, how then do I determine the x and the y?
  18. Oct 17, 2003 #17


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    Pretty much the same way- that's the whole point of calculus. Problems that are easy as long as we are working with linear functions can be extended to non-linear problems.

    Suppose the problem is, to use your example, to find the area between the x-axis (y=0) and the graph of y= sin(x) between x=0 and x= π . The upper boundary of that is y= sin(x), the lower boundary is y=0 so the integrand is sin(x)- 0= sin(x). The integral is ∫0π sin(x)dx . I remember that the derivative of cos(x) is - sin(x) so the derivative of - cos(x) is sin(x): the anti-derivative of sin(x) is - cos(x).
    ∫0π sin(x)dx= - cos(x) evaluated between π and 0. cos(π)= -1 so -cos(π)= 1 and cos(0)= 1 so -cos(0)= -1. ∫00 sin(x)dx= 1- (-1)= 2. The area between the x-axis and the graph of y= sin(x) between x=0 and x= π is 2.
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