integration area help

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  1. dongonaeatu's Avatar
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    integration area help
    find the area of the shades region; y=x^2-4

    the first area is where x is 0 and x is 2

    second area where x is 2 and x is 3

    first area is underneath x axis, second area above x axis

    so i did limits and got -16/3 for first area ( i know you just make this positive) but for the second area which is above the x axis i got -41/3? how can this be? someone help
  2. Sketch's Avatar
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    Re: integration area help
    Can i ask what the answer to the question is?
  3. dongonaeatu's Avatar
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    Re: integration area help
    (Original post by Sketch)
    Can i ask what the answer to the question is?
    well first area i got -16/3 but as its underneath x-axis you just count it as positive so 16/3. second area i got -41/3 but it is above x axis. I should add both numbers together but surely im wrong if the area above the x axis is a negative? so either 19 or -25/3 lol. what did you get can you help
  4. sputum's Avatar
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    Re: integration area help
    If you run x=3 through your integral what do you get?
  5. dongonaeatu's Avatar
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    Re: integration area help
    (Original post by sputum)
    If you run x=3 through your integral what do you get?
    -3
  6. sputum's Avatar
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    Re: integration area help
    (Original post by dongonaeatu)
    -3
    Keerect!
    Now take what you got for x=2 away from this.
  7. Sketch's Avatar
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    Re: integration area help
    hmmm, for the second area i got 7/3.. are you sure your method is correct? talk me through your working out for the second area..
  8. dongonaeatu's Avatar
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    Re: integration area help
    oh.. so is it meant to be -3--41/3? with 2 minus's making a plus
  9. dongonaeatu's Avatar
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    Re: integration area help
    (Original post by Sketch)
    hmmm, for the second area i got 7/3.. are you sure your method is correct? talk me through your working out for the second area..
    second area i did

    [(3)^3/3-4(3)] - [(2)^3/3-4(2)]= -41/3

    wait i just did it again... i got 7/3 this time

    wtf it will still end up being negative -3+7/3= -2/3

    oh wait.. the whole second area is 7/3 so its 16/3+7/3= 23/3 for the whole area?
    Last edited by dongonaeatu; 13-04-2012 at 17:12.
  10. sputum's Avatar
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    Re: integration area help
    7/3 is good
  11. Sketch's Avatar
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    Re: integration area help
    The area in this case would be the integral of the "above x axis" area, minus MINUS the integral of the "below x axis" area.. i.e. just take the sum of their absolute values..
  12. dongonaeatu's Avatar
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    Re: integration area help
    the answer for the whole area is 23/3 yes or no
  13. Sketch's Avatar
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    Re: integration area help
    where did the -3 come from may i ask?
  14. Sketch's Avatar
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    Re: integration area help
    (Original post by dongonaeatu)
    the answer for the whole area is 23/3 yes or no
    yes!
  15. sputum's Avatar
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    Re: integration area help
    (Original post by Sketch)
    where did the -3 come from may i ask?
    putting x=3 in y=x3/3 -4x
  16. dongonaeatu's Avatar
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    Re: integration area help
    Thanks for the help you two i repped you both ;D
  17. dongonaeatu's Avatar
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    Re: integration area help
    uh ran out of rep..sputnam i'll rep you later bro thanks
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