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Math Help - evaluation of an integral

  1. #1
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    evaluation of an integral

    Evaluate
    ∫ xdx + xydy where c is the line y = 1 - x for 0 ≤ x ≤ 1.
    c


    (a) By using the parameterization
    x(t) = t^2 & y(t) = 1 t^2 for 0 t 1.
    (b) By using the parameterization
    x(t) = sin t & y(t) = 1 sin t for 0 t

    2
    Note that one would normally choose
    x(t) = t & y(t) = 1t to evaluate this integral, but the above choices are selected to demonstrate that the choice of parameterization does not affect the answer.

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  2. #2
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    \mathop\int\limits_{C} xdx+xydy=\int_0^{1} t^2(2tdt)+\int_0^{1} t^2(1-t^2)(-2tdt);\quad C=\{(x,y):x=t^2,y=1-t^2\}
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  3. #3
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    Quote Originally Posted by shawsend View Post
    \mathop\int\limits_{C} xdx+xydy=\int_0^{1} t^2(2tdt)+\int_0^{1} t^2(1-t^2)(-2tdt);\quad C=\{(x,y):x=t^2,y=1-t^2\}
    thanks so much now how do i do it with x(t) = sin t and y (t) = 1 - sin t?
    i understand as far as substituting x(t) = sin t, y(t) = 1 - sin t, dx = cos t dt and dy = (- cos t) dt into the equation with limits of t = 0 to t = pi/2 but i dont know how to integrate sin t cos t or the other part? do i need to use the product rule??
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