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Math Help - Integral question

  1. #1
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    Integral question

    Consider the function .
    Let be the antiderivative of with .
    Then ?

    I'm guessing you have to find the antiderivative, then find C and then evaluate the antiderivative with 4 as your x input.

    For the antiderivative I ended up with -7/2x^2 + 6/7x^7 + 2.64

    However the webwork assignment doesn't accept my final answer of 2.42

    Can anyone please tell me where I messed up? Or at least just tell me whether my antiderivative was correct or not.
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    Or at least just tell me whether my antiderivative was correct or not.
    Looks good to me.
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  3. #3
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    Even the C value?

    Because the webwork assignment considers my answer wrong.
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  4. #4
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    I did not check the C value, you should double check that. Leave your answer as an improper fraction.
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  5. #5
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    I don't get what I'm doing wrong.

    So I plug in 4 for the x values in my antiderivative; . I get 2.42 as my answer - which is wrong.
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  6. #6
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     \int \frac{7}{x^3}-\frac{6}{x^8} ~dx = \frac{-7}{2x^{2}}+\frac{6}{7x^7} +C

     F(1)= 0 \implies 0 = \frac{-7}{2(1)^{2}}+\frac{6}{7(1)^7} +C

     0 = \frac{-7}{2}+\frac{6}{7} +C

     0 = \frac{-49}{14}+\frac{12}{14} +C

     0 = \frac{-49+12}{14}+ +C

     0 = \frac{-37}{14}+C

     C = \frac{37}{14}

    Did this rather quick, might need to check my arithmetic...

    F(x)=  \frac{-7}{2x^{2}}+\frac{6}{7x^7} +\frac{37}{14}

    F(4)=  \frac{-7}{2\times 4^{2}}+\frac{6}{7\times 4^7} +\frac{37}{14}=\dots
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  7. #7
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    That's exactly what I did. The webwork doesn't recognize it as correct. Either we both made the same mistake or the question is bugged.

    EDIT: Nevermind ... it just needed me to put the answer in 4 decimal places. Jeez, what a waste of time.

    Thanks for your help though!
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