Question Number 115267 by bobhans last updated on 24/Sep/20
$${If}\:{f}\left({x}\right)\:{is}\:{a}\:{differentiable}\:{function} \\ $$$${defined}\:\:\forall{x}\in\mathbb{R}\:{such}\:{that}\:\left({f}\left({x}\right)\right)^{\mathrm{3}} −{x}+{f}\left({x}\right)=\mathrm{0} \\ $$$${then}\:\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\:{f}^{−\mathrm{1}} \left({x}\right)\:{dx}\:=\: \\ $$
Answered by Olaf last updated on 24/Sep/20
![f^3 (t)−t+f(t) = 0 Let t = f^(−1) (x) (fof^(−1) )^3 (x)−f^(−1) (x)+fof^(−1) (x) = 0 x^3 −f^(−1) (x)+x = 0 f^(−1) (x) = x^3 +x ∫_0 ^(√2) f^(−1) (x)dx = [(x^4 /4)+(x^2 /2)]_0 ^(√2) = 1+1 = 2](Q115284.png)
$${f}^{\mathrm{3}} \left({t}\right)−{t}+{f}\left({t}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{Let}\:{t}\:=\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left({f}\mathrm{o}{f}^{−\mathrm{1}} \right)^{\mathrm{3}} \left({x}\right)−{f}^{−\mathrm{1}} \left({x}\right)+{f}\mathrm{o}{f}^{−\mathrm{1}} \left({x}\right)\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{3}} −{f}^{−\mathrm{1}} \left({x}\right)+{x}\:=\:\mathrm{0} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:{x}^{\mathrm{3}} +{x} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} {f}^{−\mathrm{1}} \left({x}\right){dx}\:=\:\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:=\:\mathrm{1}+\mathrm{1}\:=\:\mathrm{2} \\ $$