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Question Number 200105 by cortano12 last updated on 14/Nov/23

Commented by mr W last updated on 14/Nov/23

Commented by mr W last updated on 14/Nov/23

$${\int }_0^1{f}^{-1}\left(y\right)dy=1\times 1-\frac{1}{3}=\frac{2}{3}$$

Answered by witcher3 last updated on 15/Nov/23

$${\int }_0^1{\mathrm{f}}^{-}\left(\mathrm{y}\right)\mathrm{dy}$$$$\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)$$$${\int }_0^1{\mathrm{xf}}^{\prime }\left(\mathrm{x}\right)\mathrm{dx}=\left[{}_0^1\mathrm{xf}\left(\mathrm{x}\right)\right]-{\int }_0^1\mathrm{f}\left(\mathrm{x}\right)=1-\frac{1}{3}=\frac{2}{3}$$

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