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Simple integration (high school)

  1. Mar 19, 2012 #1
    zkit11.png

    For part c)

    The equation of ON is y = -0.5x which is correct;
    i know to solve it i need to find the integral between the limits 5/2 and 0 of the area of the curve - the area of the line, but what i dont understand is why that is.

    If I take the integral of the limits 2 and 0 of the curve, which area i'm i actually finding? It seems like im doing things without understanding them, could someone show me a sketch? Also, if i'm finding the integral of the line y = -0.5x of the limits 5/2 and 0, which area am i exactly finding? Why is it the integral between the two equations which is the shaded bit?

    If anyone could clear this up would be great.
     
  2. jcsd
  3. Mar 19, 2012 #2

    HallsofIvy

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    If you integrate f(x) from a to b, you get the (signed) area of the region between the graph of y= f(x) and the x-axis. The integral of [itex]2x- x^2[/itex] with respect to x from x=0 to x= 2 is the area below the curve and above the x-axis. The integral of -0.5x with respect to x from x= 0 to x= 2.5 is -1 times the area between the graph of y= -0.5 x and the x-axis.

    The graph of [itex]y= 2x-x^2[/itex] goes below the x-axis so the signed area between it and the x-axis is negative- but not "as negative" as -0.5x so the integral of [itex](2x- x^2)- (-0.5x)[/itex] is still positive.
     
  4. Mar 19, 2012 #3
    Is the area always from the graph, to the x axis? Also is this what you mean of the area (the "A" bit):
    2wncm6w.png
    If so, I still dont understand how subtraction the two would give you the area needed.
     
  5. Mar 19, 2012 #4
    The area you get from integrating like this is always measured from the x axis (if you integrate with respect to x that is). That's how the subtraction thing works. The total area under the line ON (well above actually), minus the total area above the curve from where the curve is negative to the intersection point.

    I made this image for you, it should explain it. Just remember that the area under the x-axis will be negative area!
    a9tw2.jpg
     
    Last edited: Mar 19, 2012
  6. Mar 20, 2012 #5
    So wouldnt the area we are finding by subtracting the two be this area?

    2n7kq2x.png
     
  7. Mar 20, 2012 #6
    Not quite. Figure 1 is [itex]\displaystyle\int_0^{\frac{5}{2}} (2x-x^2) \, dx[/itex] and Figure 2 is [itex]\displaystyle\int_0^{\frac{5}{2}} -\frac{1}{2}x \, dx[/itex]. Do you now see why it's correct to subtract the integrals?

    integral.gif
     
  8. Mar 20, 2012 #7
    I do now, thank you very much, can I ask what program that is you're using for the images?
     
  9. Mar 20, 2012 #8
    Glad to hear it! I used the Wolfram Alpha website to generate the graphs of the functions. I the. Saved the image to my computer, went into Microsoft Paint, drew the vertical line at 2.5 myself and filled it in myself! :)

    I'm sure there are math programs out there that can do better but that's what I did.
     
  10. Mar 21, 2012 #9
    Okay, if anyone knows any please do let me know, thanks.
     
  11. Mar 21, 2012 #10
    Okay, if anyone knows any please do let me know, thanks.
     
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