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$\int_C(x^2y^3-\sqrt{x})\,\mathrm{d}y$, $C$ is the arc of the curve $y=\sqrt{x}$ from $(1,1)$ to $(4,2)$

I set $y=t$ and therefore $x=t^2$ on the interval as $t$ goes from $1$ to $2$ and also $\mathrm{d}y=\mathrm{d}t$. Plugging in and solving I get $\int_1^2t^6-t\,\mathrm{d}t$. Unfortunately I'm not getting the right answer, what am I doing wrong?

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  • $\begingroup$ Just asking, since I'm solving a similar problem right now — how is it possible to set $y=t$? EDIT: Never mind, because $dy=\frac{dy}{dt}dt$. $\endgroup$ – user3932000 May 12 '16 at 14:18
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$x^2y^3=(t^2)^2t^3=t^4t^3=t^7$.

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  • $\begingroup$ Oops! I wrote down $y^2$ on my paper. $\endgroup$ – ShaneBird Nov 13 '14 at 18:35

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