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Math Help - Integration :(

  1. #1
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    Integration :(

    Im doing q's were i need to find the area of curves on a graph but im going wrong somewhere. please help

    y=x^5-2 between x=-1 and x=0 if differentiated to 1/6x^6 - 2x
    and drawn the graph of x^5-2

    then by substituting x to be o and -1
    {1/6(-1)^6 - 2(-1)}-{1/6(0)^6-2(0)]
    and got the answer to be25/6 but the answer in the book is 2 1/6
    thank you
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  2. #2
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    Quote Originally Posted by Chez_ View Post
    Im doing q's were i need to find the area of curves on a graph but im going wrong somewhere. please help

    y=x^5-2 between x=-1 and x=0 if differentiated to 1/6x^6 - 2x
    and drawn the graph of x^5-2

    then by substituting x to be o and -1
    {1/6(-1)^6 - 2(-1)}-{1/6(0)^6-2(0)]
    and got the answer to be25/6 but the answer in the book is 2 1/6
    thank you

    Your post is a bit muddled but all you need to do is work out

    \frac{1}{6}(-1)^6-2\times-1

    which you already had correct and it does give the answer you quoted.

    (-1)^6 = 1 and -2 x -1 =2

    so you have 1/6+2= 2 1/6

    By the way you normally sub in the upper limit first so in this case the area is negative (below the x axis) but the answer you should give is the absolute value as given in the book 2 1/6.
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