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Math Help - Fundamental Theorem of Calculus to evaluate the integrals

  1. #1
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    Fundamental Theorem of Calculus to evaluate the integrals

    The integral sign has a 2 at the top and a 0 at the bottom

    2/x^2+1 dx
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    Quote Originally Posted by cara View Post
    The integral sign has a 2 at the top and a 0 at the bottom

    2/x^2+1 dx
    \int_0^2 \frac{2}{x^2} + 1 \, dx or \int_0^2 \frac{2}{x^2+1} \, dx ???
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    The second one you listed is the problem that I need
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    Quote Originally Posted by cara View Post
    The second one you listed is the problem that I need
    You have \int_{0}^{2}\dfrac{2}{x^2+1}\;{dx}. Let x = \tan\theta, then \dfrac{dx}{d\theta} = \sec^2\theta \Rightarrow dx =  \left(d\theta\right){\sec^2\theta}. When x = 2, \tan\theta = 2 \Rightarrow \theta = \arctan\left(2\right) . When x = 0, \tan{\theta} = 0 \Rightarrow \theta = 0 . So we have \int_{x=0}^{x=2}\dfrac{2}{x^2+1}\;{dx} = \int_{\theta=0}^{\theta=\arctan(2)}\dfrac{2}{\tan^  2\theta+1}{\sec^2\theta}\;{d\theta} = \int_{0}^{\arctan(2)}{2}\;{d\theta} = \bigg[2\theta\bigg]_{0}^{\arctan(2)} = \boxed{2\arctan(2)}.
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    we are not using arc tan, but F(b)-F(a)
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    Quote Originally Posted by cara View Post
    we are not using arc tan, but F(b)-F(a)
    uhh ...

    F(2) - F(0) = 2[\arctan(2) - \arctan(0)] = 2\arctan(2)

    ... as stated previously
    Last edited by skeeter; April 22nd 2010 at 05:29 PM. Reason: typo
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