(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]

\int_{C}(xy+\ln{x})\mathrm{d}s

[/tex]

where C is the arc of the parabola [itex]y=x^2[/itex] from (1,1) to (3,9)

2. Relevant equations

[tex]

\int_{C}f(x,y)\mathrm{d}s = \int_{a}^{b}f(x(t),y(t))\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\mathrm{d}t

[/tex]

3. The attempt at a solution

Ok, so I take x as the parameter t, so the parametric equations of the line [itex]x=t[/itex] and [itex]y=t^2[/itex] where [itex]t\in [1,3][/itex]. I can then rewrite this line integral as a regular definite integral by using the formula mentioned above (pardon the bad formatting). However, when I do that, I come up with

[tex]

\int_{1}^{3}(t^3+\ln{t})\sqrt{1+4t^2}\mathrm{d}t

[/tex]

I can't seem to find a way to integrate this function. Itisintegrable, at least according to the Wolfram Integrator. A solutions manual for the book I'm studying from comes up with

[tex]

\int_{1}^{3}(t^3+\ln{t})\sqrt{4t^2}\mathrm{d}t = \int_{1}^{3}(t^3+\ln{t})2t\mathrm{d}t

[/tex]

which is quite straightforward, however, it seems to me that it's a mistake since [itex]\frac{dx}{dt}=1[/itex] thus there is a 1 missing under the square root.

So can someone please help me find a way to integrate the integral I've come up with or am I missing something simple, i.e. ithasbeen set up correctly in the book?

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# Homework Help: A cumbersome line integral

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