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C is the arc of the curve $y=\sqrt{x}$ from $(1,1)$ to $(4,2)$. Find $$\int_cx^2y^3-\sqrt{x}\space\mathrm{d}y$$Looks simple enough. I take $x=t$ and $y=\sqrt{t}$. This leaves $$\int_1^2[t^2\cdot t^\frac{3}{2}-\sqrt{t}]\frac{1}{2\sqrt{t}}\mathrm{d}t=\frac{11}{8}$$ Sadly, the answer is actually $\frac{243}{8}$.

What is the error in my logic?

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The $2$ is supposed to be a $4$... $y=2$ hence $t=4$, so the upper bound is $4$. –  Shahar Jul 21 at 1:05

1 Answer 1

Everything is correct except the limits of integration. The curve is from $(1,1)$ to $(4,2)$, so $t$ ranges from $1$ to $4$. Change the limits, and you'll get $\frac{243}{8}$.

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