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Integration, area under line and curve.

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data

    The diagram shows the graph of the curve C whose equation is [tex] y = x(x-1)(x-2) [/tex]

    a) find the equation of the tangent at the pont P whose x coordinate is [tex] \frac{1}{2} [/tex]

    b) show that the point Q(2,0) lies on the tnagent and on C.

    c) Find the shaded area.


    the equation of the tangent line is [tex] y = -\frac{1}{4}x +\frac{1}{2} [/tex]

    But I am not sure how to work out the shaded area?

    Is it the line minus the x-axis?

    [tex] \int_\frac{1}{2}^2 -\frac{1}{4}x + \frac{1}{2} -( \frac{3}{2}) [/tex] [tex] dx [/tex]

    ?

    https://www.physicsforums.com/attachment.php?attachmentid=18690&d=1241015314
     

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    Last edited: Apr 29, 2009
  2. jcsd
  3. Apr 29, 2009 #2

    Tom Mattson

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    Draw rectangles of differential area. What curve does the top of the rectangle touch? What curve does the bottom touch? Hint: For the latter question the answer is different on [itex][\frac{1}{2},1)[/itex] than it is on [itex](1,2][/itex].
     
  4. Apr 29, 2009 #3

    The height of the rectanlge touches the tangent line, and the bottom touches the x-axis?

    Why are you also including the area under the x-axis , which is not shaded?
     
    Last edited: Apr 29, 2009
  5. Apr 29, 2009 #4
    Basically what it wants you to do here is like you said work out the area of the triangle PQY. Then use definite integration between P & 1 (when the curve goes below). I don't think you need the coordinates of Q at all here.

    Can you follow it from there?

    if not ask
     
    Last edited: Apr 29, 2009
  6. Apr 29, 2009 #5

    Do you mean 'PQX'?
     
  7. Apr 29, 2009 #6
    sorry, yes =p x being ( 0.5 , 0 ) :D
     
    Last edited: Apr 29, 2009
  8. Apr 29, 2009 #7

    I dont think that makes a triangle though.
     
  9. Apr 29, 2009 #8
    Hm? My fault, perhaps i'm not being very clear.

    You take P right? Draw a line straight down to the X-axis, then follow it across to Q. This will form a right angled triangle. I think you can follow my train of thought here, if you can work out the area of that triangle...think about how you can work out the area of the shaded region using definite integration!

    tip;
    [tex]\int ^{1} _{0.5}[/tex]

    =D
     
  10. Apr 29, 2009 #9
    Oh right,

    so your saying find the area of that triangle then find the area under the curve using definite integration from 0.5 to 1 and minus this area away from the triangle ?
     
  11. Apr 29, 2009 #10
  12. Apr 29, 2009 #11
    Okay I'll try that and see what I get!

    thanks.
     
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