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Math Help - Another Calculus Problem.

  1. #1
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    Another Calculus Problem.

    e
    ∫ (x^2-1)/(x) dx=
    1


    A) e- (1/e)

    B) e^2 - e

    C) (e^2/2)-e+ (1/2)

    D) e^2-2

    E) (e^2/2)-(3/2)
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  2. #2
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    Quote Originally Posted by frozenflames
    e
    ∫ (x^2-1)/(x) dx=
    1


    A) e- (1/e)

    B) e^2 - e

    C) (e^2/2)-e+ (1/2)

    D) e^2-2

    E) (e^2/2)-(3/2)
    First find the antiderivative of
    \int\frac{x^2-1}{x}dx
    Which is,
    \int x-\frac{1}{x}dx
    Which is,
    \frac{x^2}{2}-\ln |x|,
    But it is taken from [1,e]
    Which by the fundamental theorem is,
    \frac{e^2}{2}-1-\frac{1}{2}=\frac{e^2}{2}-3/2
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