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Math Help - Integration

  1. #1
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    Integration

    I'm a bit confused about this

    Let G(x)=\int_{0}^{x} f(x)dx

    The graph of f(x) is under the x axis from [-2,0]

    Wouldn't the graph of G(x) be increasing on the interval of [-2,0] because

    G(x)=\int_{0}^{-2} f(x)dx=-\int_{-2}^{0} f(x) dx = -(-area)=positive area

    But apparently the graph is decreasing on that interval. Can someone explain this to me? Thanks
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  2. #2
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    Quote Originally Posted by Linnus View Post
    I'm a bit confused about this

    Let G(x)=\int_{0}^{x} f(x)dx

    The graph of f(x) is under the x axis from [-2,0]

    Wouldn't the graph of G(x) be increasing on the interval of [-2,0] because

    G(x)=\int_{0}^{-2} f(x)dx=-\int_{-2}^{0} f(x) dx = -(-area)=positive area Mr F says: This is G(-2), NOT G(x).

    But apparently the graph is decreasing on that interval. Can someone explain this to me? Thanks
    Correct, it is indeed decreasing over the interval -2 \leq x \leq 0.

    Consider:

    G(-2) = \int_{0}^{-2} f(x) \, dx = - \int^{0}_{-2} f(x) \, dx > 0.

    G(-1) = \int_{0}^{-1} f(x) \, dx = - \int^{0}_{-1} f(x) \, dx > 0.

    But clearly G(-2) > G(-1) .......
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Correct, it is indeed decreasing over the interval -2 \leq x \leq 0.

    Consider:

    G(-2) = \int_{0}^{-2} f(x) \, dx = - \int^{0}_{-2} f(x) \, dx > 0.

    G(-1) = \int_{0}^{-1} f(x) \, dx = - \int^{0}_{-1} f(x) \, dx > 0.

    But clearly G(-2) > G(-1) .......
    Ah...that makes sense.
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