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Understanding the area under a curv and trapizum rule

  1. Jan 18, 2013 #1
    I am having a bit of difficulty understanding the intuition behind it. So here my understanding so far, I have attached some photos of the book I am using to understand integration.

    This is from the book I have posted, I am just going up to the part I do not understand.

    S basically I have to consider a rectangle ( which is the one at the bottom of the first photograph ). This rectangle has a width of δx. An area of δA which is the shaded area.

    The area of rectangle ABEF is yδx: which is pretty self explanatory.

    The area of rectangle ABCD is (y+δy)δx

    Therfore we have: yδx<δA <(y+δy)δx: So all this is doing is comparing the area to the 2 rectangles

    Now the confusion begins, it say beceause δx >0 You should divided through by. Why is this?

    Letting δx [itex]\rightarrow[/itex]0 gives [itex]\frac{δA}{δx}\rightarrow\frac{dA}{d} [/itex] and δy[itex]\rightarrow[/itex]0 therefore [itex]\frac{dA}{dx}=y[/itex] this to me is saying that x is the independent variable and δA dependent variable; correct? So as more rectangle are add then obviously the area will get smaller and the more rectangles add δy will reach zero. But how dose [itex]\frac{dA}{dx}=y[/itex]

    This is the trouble I am having up to a point, if someone could ans and the I will ask the other half of the trouble I am having. Or if someone feel like it explain from the beginning how the intergration of the curve works.

    I am new to integration so I would appreciate all the help anyone can offer.
     
  2. jcsd
  3. Jan 18, 2013 #2
    Did you mean to include some images with your post? Because you didn't.
     
  4. Jan 18, 2013 #3
    Oh yeah that would help, give me sec they will be uploaded
     
  5. Jan 18, 2013 #4
    Here they are, sorry about that
     

    Attached Files:

  6. Jan 18, 2013 #5

    Mark44

    Staff: Mentor

    The ideas are pretty simple, actually. You have a typical area element that is the exact area under the curve between xi and xi+1 = Δx. You can approximate this area by a rectangle or by a trapezoid or by some other geometric figure.

    If you approximate the area by a rectangle there are a couple of possibilities:
    1. Use the left endpoint as the height of the rectangle, getting an area of f(xi)* Δx.
    2. Use the right endpoint as the height of the rectangle, getting an area of f(xi + 1)* Δx.

    If you approximate the area by a trapezoid instead of a rectangle, the top side of the trapezoid is a straight line between (xi, f(xi)) and (xi+1, f(xi+1)). The trazezoid approximation uses the idea that the area of a trapezoid is the base times the average of the two unequal sides.

    IOW, the approximate area is (1/2)[ f(xi+1) + f(xi+1)] * Δx.

    That's really all there is to it.
    That would be Δx.
    That would be ΔA.
    That would be ##\frac{ΔA}{Δx}##, which does NOT have a limit of dA/d. dA/d is meaningless. Anyway, why are you taking this limit? To get the area under the curve, you want to add all the ΔA pieces, and that's what the Riemann sum is about.
    ??
     
  7. Jan 18, 2013 #6
    If you have an inequality, then multiplying or dividing both sides by a positive number keeps the inequality sign the same, while if you multiply or divide both sides by a negative number the inequality sign gets flipped. For instance, take the inequality 3<6. If you divide both sides by 3, you get 1<2. But if you instead divided both sides by -3, then you would get -1>-2; the less than sign changes to a greater than sign. So since δx is a positive number, when we divide by it the inequality signs stay the same.
    You mean [itex]\frac{δA}{δx}\rightarrow\frac{dA}{dx}[/itex].
    Yes, δA is a variable that depends on both x and δx. δy is also a variable that depends on both x and δx.
    Yes, as the rectangles become thinner and thinner, the amount δy that the function changes by between x and x+δx goes to zero.
    Let me explain that. We have the inequality y < δA/δx < y+δy. Now as δx goes to 0, y doesn't change, but δy goes to zero, so y+δy → y. Now there's a theorem in calculus, known as the squeeze theorem, that says that if f(x)<g(x)<h(x), and the limit of f(x) as x goes to a equals L, and the limit of h(x) as x goes to a equals L, then that implies that the limit of g(x) as x goes to a is also equal to L. So the limit of y as δx goes to 0 equals y, and the limit of y+δy as δx goes to 0 equals y, therefore by the squeeze theorem the limit of δA/δx as δx goes to 0 is also equal to y. And then we use the fact that the limit of δA/δx as δx goes to 0 is equal to dA/dx, therefore dA/dx = y.
    What's the other half of your trouble?
    What exactly are you asking?
     
  8. Jan 20, 2013 #7
    Right, first of all thanks for the replies. The have been a big help. I see how the rule work, but really its only and approximation, not a definite integral, because not matter how many rectangles or trap you will always have gaps in-between the curve. I think this is the real part that causing the confusion. I watch a couple of vids and read more articles but, nothing really ans my question.

    Once again big for any advice that anyone could give.
     
  9. Jan 20, 2013 #8
    Yes, but as you add more and more rectangles or trapezia, the size of those gaps gets smaller and smaller. That's the point of approximation methods: you can usually make them as accurate as you wish, as long you're willing to do the extra computations.
     
  10. Jan 20, 2013 #9
    So for example if, I workout the area underneath a curve of f(x) between two point a and b and take the integral. I am really getting a very close approximation to the actuall area under the curve. So when you say If your willing to do the calculation, you mean your are referring to the amount of ordinates, under a given curve?
     
  11. Jan 20, 2013 #10
    Yes, for any finite number of rectangles, it's only an approximation. But as the number of rectangles becomes bigger and bigger, the gaps become smaller and smaller, so the approximation becomes better and better. So in the limit as the number of rectangles goes to infinity, the thickness of each rectangle goes to zero, the size of the gaps goes to zero, and thus the approximate area becomes the exact area. Do you understand that point?
     
  12. Jan 20, 2013 #11
    The integral is exactly the area the curve, not just very close. It is the limit of these approximation methods as the number of rectangles is taken to infinity. For a finite number of rectangles, the result is approximate.
     
  13. Jan 20, 2013 #12
    Right okay I have clocked it now. Thanks for clearing that, up for me. Some of my confusion stems from the book I read, that don't really show how, the integral dose what is dose more of here how to workout the math. If you catch my drift.
     
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