int(ydx + xdy); C is the curve y=x^2, 0 <= x <= 1 Thanks
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Originally Posted by pakman int(ydx + xdy); C is the curve y=x^2, 0 <= x <= 1 Thanks I have little time (might be wrong). It seems that, f(x,y)=xy Is a scalar potential. Thus, by the fundamental theorem of line integrals, f(1,1)-f(0,0)=1
Originally Posted by ThePerfectHacker I have little time (might be wrong). It seems that, f(x,y)=xy Is a scalar potential. Thus, by the fundamental theorem of line integrals, f(1,1)-f(0,0)=1 I figured it out actually. Since y is not changing, dy = 0 thereby removing the xdy part of the function. int(y dx) is all that remains. Plug in x^2. int(x^2 dx) which equals 2 after integrating from 0 to 1.
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