Area Calculation for Parametric Equation: x=t^3-5t, y=7t^2

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Homework Statement


Find the area of the region enclosed by the parametric equation
x=t^3-5t
y=7t^2


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The Attempt at a Solution


I know how you set it up \int (7t^2)(3t^2-5)dt, but how do you find the bounds. I tried finding t and got t= (+/-)\sqrt{y/7} and you plug it into x but where do you go from there.
 
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What I understand by enclosed means for the value of t where x=0. So the values are x=-\sqrt{5} , 0 , \sqrt{5} So I guess you have to find two integrals, for x>0 and for x<0 because one will get negative and the other positive and to add them.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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