MHB 16.1.9 Line Integral over space curves

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To evaluate the line integral $\int_C(x+y)ds$ over the straight-line segment defined by $x=t$, $y=(1-t)$, and $z=0$, where the path C runs from the point (0,1,0) to (1,0,0), first express $ds$ using the formula $ds = \sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}dt$. The derivatives are calculated as $dx/dt = 1$, $dy/dt = -1$, and $dz/dt = 0$. The integral is then set up with limits for $t$ from 0 to 1, substituting the expressions for $x$ and $y$ into the integral. This approach simplifies the evaluation of the integral, leading to the final result.
karush
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Evaluate

$\displaystyle \int_C(x+y)ds$
where C is the straight-line segment
$x=t, y=(1-t), z=0, $
from (0,1,0) to (1,0,0)

ok this is due tuesday but i missed the lecture on it
so kinda clueless.
i am sure it is a easy one.
 
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Use $ds = \sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}dt$, take derivative of $(x(t),y(t),z(t))$ with respect to t, and then the integral goes from $(x,y,z)=(0,1,0) $ to $(1,0,0)$ check what it means for the t variable.