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Find the value of $\int{F \cdot \mathrm{d}r}$, where $$F(x,y) = \langle 5e^y+ye^x,e^x+5xe^y \rangle$$ and $$C: r(t) = \left\langle\sin\left(\frac{\pi t}{2}\right),\ln(t)\right\rangle; 1\le t\le2$$ So far, I've tried substituting in using the parametriziation suggested by the $r(t)$ function into the $F(x,y)$ vector field but that produced such a big integral there's no way of evaluating it. Any help would be welcome, thank you in advance!

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  • $\begingroup$ If \begin{align} r &= \langle x(t), y(t) \rangle \\ &= \langle \sin(at), \ln(t) \rangle \end{align} then \begin{align} dr &= \langle dx, dy \rangle \\ &= \langle a \cos(at) dt, dt/t \rangle \end{align} and \begin{align} F &= \langle 5e^y + ye^x, e^x + 5xe^x \rangle \\ &= \langle 5t + \ln(t) e^{\sin(at)}, e^{\sin(at)} + 5 \sin(at) e^{\sin(at)} \rangle \end{align} and dotting gives $$\int_{1}^{2} [5t + \ln(t) e^{\sin(at)}] a \cos(at) + \frac{e^{\sin(at)} + 5 \sin(at) e^{\sin(at)}}{t} dt $$ To solve, use integration by parts and $u$ substitutions. $\endgroup$ – Mattos Mar 21 at 4:06
  • $\begingroup$ Well, for one, integration by parts on $\ln(t) \cdot a \cos(at) e^{\sin(at)}$, with $u = \ln(t) \implies u' = 1/t$ and $v' = a \cos(at) e^{\sin(at)} \implies v = e^{\sin(at)}$ gives $$I = \ln(t) e^{\sin(at)} \bigg \lvert_{1}^{2} - \int_{1}^{2} \frac{e^{\sin(at)}}{t} dt$$ where the integral part of $I$ cancels with the $$\color{red} + \int_{1}^{2} \frac{e^{\sin(at)}}{t} dt$$ from our original integral. $\endgroup$ – Mattos Mar 21 at 4:18
  • $\begingroup$ @Mattos This should be an answer, not a comment. $\endgroup$ – Naman Kumar Mar 21 at 4:52
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As seen in the comments (and corrected on one key point), we can parametrize and set up the integral $$\int_C F\cdot d\mathbf{r} = \int_1^2 [5t+\ln(t)e^{\sin(at)}]a\cos at + \frac{e^{\sin(at)}+5\sin(at)e^{\ln t}}{t}\,dt$$ where $a=\frac{\pi}{2}$, abbreviated to simplify the typography.

This works, but it's messy.

Instead, note that we can write $F$ as a gradient $F=\nabla G$, where $G(x,y)=5xe^y+ye^x$. Then $G$ acts as an antiderivative; we can apply the line integral version of the FTC to get $$\int_C F\cdot d\mathbf{r} = G(r(2))-G(r(1)) = G(0,\ln 2) - G(1,0) = \ln 2 - 5$$

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  • $\begingroup$ The answer isn't 0 though. (thank you for your help) $\endgroup$ – James Done Mar 21 at 4:26
  • $\begingroup$ Corrected. I saw one term zero at each endpoint, and interpreted it as both terms zero as each endpoint. Not anymore. $\endgroup$ – jmerry Mar 21 at 4:29
  • $\begingroup$ Great, thank you, that's a beautiful solution. $\endgroup$ – James Done Mar 21 at 4:30

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