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How can I find the length of the curve $$\left(\frac{t^3}{3} - t\right)\mathbf{i}+ t^2 \mathbf{j}, \quad 0≤t≤1?$$

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When we do the calculation, if we do not make a minor slip, the thing inside the square root turns out, "miraculously," to be a perfect square. Change the numbers a little and you will get something you can't integrate in elementary terms. –  André Nicolas Oct 30 '12 at 18:49

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Length of a curve from $t=a$ and $t=b$ is given by $$\int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} dt$$ provided $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ exists and are continuous.

In your problem, $x(t) = \dfrac{t^3}3 - t$, $y(t) = t^2$, $a = 0$ and $b = 1$.

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@BabakSorouh True. I will update it. –  user17762 Oct 30 '12 at 18:48
    
thank you so much for your answer. this helped me to much. –  Yigit Oct 30 '12 at 18:50

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