How can I prove $$\int_0^1 \sqrt{1-x^3}dx=\int_0^1 \sqrt[3]{1-x^2}dx$$
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possible duplicate of Is $\int_a^b f(x) dx = \int_{f(a)}^{f(b)} f^{-1}(x) dy$? – Mar 15 '15 at 00:26
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Hint: If $f(x) = \sqrt{1-x^3} $, then $ f^{-1} (x) = \sqrt[3]{1-x^2}$
Observe that $ f(0)=1, f(1)=0$.
Calvin Lin
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The first integral describes the area enclosed in the first quadrant by the curve $$x^3 + y^2 = 1,$$ and the second describes the area enclosed in the first quadrant by the curve $$x^2 + y^3 = 1.$$ So with the transformation $(x,y) \to (y,x)$, it is obvious the two areas are equal. Alternatively, we can regard both integrals as their respective double integrals: $$\int_{x=0}^1 \int_{y=0}^{\sqrt{1-x^3}} 1 \, dy \, dx = \int_{y=0}^1 \int_{x=0}^{\sqrt[3]{1-y^2}} 1 \, dx \, dy.$$
heropup
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If we take the derivative of integrals, we will get $$\sqrt{1-x^3}=\sqrt[3]{1-x^2}$$ It is very clear the two formula are equal at $0$,$1$
E.H.E
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This doesn't show anything. $f(x)=0$ and $f(x)=x^4-x^2$ have identical values at $\pm 1$ but $\int_{-1}^10\ dx\neq\int_{-1}^1x^4-x^2\ dx$... – Steven Stadnicki Mar 14 '15 at 23:33