- #1
mmh37
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I am still struggling my way through line integrals, and this here is one where I do not understand what has gone wrong - does anyone see what it is ( I really want to understand all of this)?
[tex] \int_{0}^{1}{(y*e^{xy} + 2x + y)dx + (x*e^{xy} + x )dy } [/tex]
the curve joins (0,0) to (1,1)
where x= t and y=t (0<= t <= 1)
so, I said that
[tex] \frac {dx} {dt} = \frac {dy} {dt} = 1 [/tex]
therefore:
[tex] \int_{0}^{1} {( t*e^{t^2} + 2t + t + t*e^t^2 + t)dt} = \int_{0}^{1}{ (2t*e^{t^2} + 4t) dt} = e^1 + 1 [/tex]
However, the solution is supposed to be -2pi
[tex] \int_{0}^{1}{(y*e^{xy} + 2x + y)dx + (x*e^{xy} + x )dy } [/tex]
the curve joins (0,0) to (1,1)
where x= t and y=t (0<= t <= 1)
so, I said that
[tex] \frac {dx} {dt} = \frac {dy} {dt} = 1 [/tex]
therefore:
[tex] \int_{0}^{1} {( t*e^{t^2} + 2t + t + t*e^t^2 + t)dt} = \int_{0}^{1}{ (2t*e^{t^2} + 4t) dt} = e^1 + 1 [/tex]
However, the solution is supposed to be -2pi
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