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I have to find the line integral of the following.

$$\int_Cxe^ydx+x^2ydy, C: 0 \leq x \leq 2, y = 3$$

I am trying to understand the concept of line integrals, but in this case, I am confused as to what $C$ really is. Parameterizing both $x$ and $y$, i have $x = t$ and $y = 3$. Following from this, we have:

$$\int_0^2te^3dt+\int_0^23t^2dt$$

$$(2e^3-0)+(2^3-0)$$

$$2e^3+8$$

However, upon checking with my answer key, the answer is $2e^3$. So I am rather stuck here.

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    $\begingroup$ If $y =$ constant then $dy = 0$ $\endgroup$ – jim Apr 14 '16 at 22:32
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Your curve will be $r(t) = (t,3)$, for $0 \leq t \leq 2$. We can think that if $r(t)=(x(t),y(t))$, then ${\rm d}x = x'(t)\,{\rm d}t$ and ${\rm d}y = y'(t)\,{\rm d}t$, so here we'll have ${\rm d}x = {\rm d}t$ and ${\rm d}y = 0$. So: $$\int_C xe^y\,{\rm d}x + x^2y\,{\rm d}y = \int_0^2 te^3\,{\rm d}t = 2e^3.$$

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