Find a formula for a constant function using the mean value theorem

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Homework Statement


Let x ϵ R such that f'(x) = 3x^2. Prove that f(x) = x^3 + c for some c ϵ R using the Mean Value Theorem.


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The Attempt at a Solution


I used two functions f(x) and g(x) that have the same derivative namely f'(x). Applying the theorem I am able to work up until f(x) = g(x), I am unsure of where to go from there.
 
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If you reached the conclusion f(x) = g(x) from only the hypothesis f'(x) = g'(x), something is clearly wrong, since that deduction is false.

You have a function that's expected to be a constant, namely f(x) - x^3. Work with that.
 
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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