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Area bounded by two curves

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data

    find the Area Bounded by the two curves, y=|x+1|, y= - ( x+1)2 + 6

    2. Relevant equations

    y=|x+1|, y= - ( x+1)2 + 6

    3. The attempt at a solution


    A= Integration of | f (x) - g(x) |

    x+1= f(x)
    -(x+1)2 + 6= g(x)


    getting the limit of integration:

    x+1= - (x+1)2 + 6

    x2 + 3x - 4=0

    (X+4) ( x-1)

    so x=-4, and x=1

    -----------------------------------------

    now when dividing the absolute value


    y= x+1 ; x<-1
    y=-(x+1) ; x>-1


    Integration is denoted by { ( upper limit , lower limit ) |f(x)|

    so the Area= {(-1,-4) |(-x-1) + (x+1)2 -6 | - {(-1,1) |(x+1)+(x+1)2 - 6|








    what's wrong in this solution, I think that we should use -x-1 to get the limits as well so the area will be (the integration from -4 to -1) - (the integration from -1 to 3)
     
    Last edited: Apr 25, 2010
  2. jcsd
  3. Apr 25, 2010 #2

    tiny-tim

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    welcome to pf!

    hi the white sou!
    No … you need to use |x+1|, not x+1. :wink:
     
  4. Apr 25, 2010 #3
    Graph it first.

    so the Area= {(-1,-4) |(-x-1) + (x+1)2 -6 | - {(-1,1) |(x+1)+(x+1)2 - 6|

    Looks like you are using -4 as one of your limit however,
    at -4
    y=|x+1|, y= - ( x+1)2 + 6
    y = 3, y = -3
    They don't intersect at -4

    (tiny-tim already pointed out)
     
  5. Apr 25, 2010 #4
    Re: welcome to pf!

    okay |x+1| consists of x+1, -x-1


    once we use x+1=-(x+1)2 +6


    x= 1, x= -4

    and when we are using -x-1= -(x+1)2 + 6

    we get x=-2, x=3

    now we are having 4 points of intersection


    soo how would we get the limits
     
  6. Apr 25, 2010 #5



    I already graphed it


    for |x+1| there will be two lines intersecting at the point -1 and for the other curve we are having the the head of the curve at (-1,6) and it is opening downword
     
  7. Apr 25, 2010 #6
    When you used x+1, you got x= 1, x= -4
    When you used -(x+1), you got x=-2, x=3

    Does x+1 is right or left line in the curve |x+1|. Label |x+1| with x+1 and -(x+1) and you should be able to identify which two of the above four solutions is valid
     
  8. Apr 25, 2010 #7

    tiny-tim

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    An alternative method:

    both curves are obviously symmetric about the line x + 1 = 0, ie x = -1,

    so the solutions must also be symmetric about x = -1. :wink:
     
  9. Apr 25, 2010 #8


    so Area= {(-2,-1) |f(x)-g(x)| - {(-1,1) |f(x)-g(x)|


    am i right now?
     
  10. Apr 25, 2010 #9
    1)
    Should have been
    x= 2, x=-3
     
  11. Apr 26, 2010 #10
    okay now I knew my mistake


    thank you bro :)
     
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