# Thread: evaluation of an integral

1. ## 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.

2. $\mathop\int\limits_{C} xdx+xydy=\int_0^{1} t^2(2tdt)+\int_0^{1} t^2(1-t^2)(-2tdt);\quad C=\{(x,y):x=t^2,y=1-t^2\}$

3. Originally Posted by shawsend
$\mathop\int\limits_{C} xdx+xydy=\int_0^{1} t^2(2tdt)+\int_0^{1} t^2(1-t^2)(-2tdt);\quad C=\{(x,y):x=t^2,y=1-t^2\}$
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??