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Math Help - Improper Integral

  1. #1
    Member WhoCares357's Avatar
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    Improper Integral

    \int^3_{-\infty}\frac{dx}{x^2+9}
    I got to the part where \lim_{b->-\infty}\frac{1}{3}tan^{-1}(1)-\frac{1}{3}tan^{-1}(\frac{b}{3})
    I'm not sure how to evaluate arctan to -infinity...

    - Wolfram|Alpha[1%2F(x^2%2B9)%2C[x]]
    - Wolfram|Alpha[1%2F%28x^2%2B9%29%2C[x%2C-infinity%2C3]]
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  2. #2
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    Quote Originally Posted by WhoCares357 View Post
    \int ^3_{-\infty}\frac{dx}{x^2+9)
    I got to the part where \lim_{b->-\infty}\frac{1}{3}tan^{-1}(1)-\frac{1}{3}tan^{-1}(\frac{b}{3})
    I'm not sure how to evaluate arctan to -infinity...

    - Wolfram|Alpha[1%2F(x^2%2B9)%2C[x]]
    - Wolfram|Alpha[1%2F%28x^2%2B9%29%2C[x%2C-infinity%2C3]]
    familiar with the graph of the arctangent function?

    \lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}
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  3. #3
    Member WhoCares357's Avatar
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    Quote Originally Posted by skeeter View Post
    familiar with the graph of the arctangent function?

    \lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}
    Thanks.
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