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Math Help - integral help

  1. #1
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    integral help

    How do I find the integral of

    F(x) = the integral of 1/1+t^2 dt from 0 to x + the integral of 1/1+t^2 dt from 0 to 1/x?

    and how do I determine that F(x) is constant on (-infinity,0) and constant on (0,infinity)?

    Any help appreciated!

    Thanks
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  2. #2
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    Quote Originally Posted by amma0913 View Post
    How do I find the integral of

    F(x) = the integral of 1/1+t^2 dt from 0 to x + the integral of 1/1+t^2 dt from 0 to 1/x?

    and how do I determine that F(x) is constant on (-infinity,0) and constant on (0,infinity)?

    Any help appreciated!

    Thanks
    You should know that \int \frac{1}{1 + t^2} \, dt = \tan^{-1} (t) + C. Apply this standard result and simplify (using a right triangle might be helpful ....)
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  3. #3
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    Quote Originally Posted by amma0913 View Post
    How do I find the integral of

    F(x) = the integral of 1/1+t^2 dt from 0 to x + the integral of 1/1+t^2 dt from 0 to 1/x?

    and how do I determine that F(x) is constant on (-infinity,0) and constant on (0,infinity)?

    Any help appreciated!

    Thanks

    F(x)=\int\limits_0^x\frac{dt}{1+t^2}+\int\limits_0  ^{1\slash x}\frac{dt}{1+t^2}=\left[\arctan t\right]_0^x+\left[\arctan t\right]_0^{1\slash x}=\arctan x+\arctan\left(1\slash x\right).

    To determine that this function is a constant in some interval just differentiate it...what should you get?

    Tonio
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  4. #4
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    Since (F+ G)'= F+ G', you could also show this function is a constant by differentiating each of the integrals separately and adding. By the "Fundamental Theorem of Calculus" the derivative of the first integral is just \frac{1}{1+ x^2}. By the "Fundamental Theorem of Calculus", together with the chain rule, the derivative of the second integral is \frac{1}{1+ \frac{1}{x^2}}\left(\frac{-1}{x^2}\right)= -\frac{1}{1+ x^2}. Those clearly add to 0, for all x (except x= 0 where \frac{1}{1+ \frac{1}{x^2}} is not defined), so the function is a constant.
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