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Question Number 134947 by bobhans last updated on 08/Mar/21

$${\int }_0^2(\sqrt{1+{\mathrm{x}}^3}+\sqrt[3]{{\mathrm{x}}^2+2\mathrm{x}})\mathrm{dx}?$$

Answered by EDWIN88 last updated on 09/Mar/21

$$\mathrm{I}={\int }_0^2(\sqrt{1+{\mathrm{x}}^3}+\sqrt[3]{{\mathrm{x}}^2+2\mathrm{x}})\mathrm{dx}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y}=\sqrt[3]{{\mathrm{x}}^2+2\mathrm{x}}\mathrm{for}\mathrm{x}\in \mathbb{R}\mathrm{and}\mathrm{x}⩽-2\mathrm{or}\mathrm{x}⩾0$$$${(\mathrm{x}+1)}^2={\mathrm{y}}^3+1,\mathrm{x}+1=\sqrt{{\mathrm{y}}^3+1}$$$$\mathrm{x}=\sqrt{{\mathrm{y}}^3+1}-1,\mathrm{replace}\mathrm{x}\mathrm{by}\mathrm{y}\mathrm{and}\mathrm{we}\mathrm{get}$$$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^3+1}-1$$$$\mathrm{so}\mathrm{the}\mathrm{integral}\mathrm{becomes}$$$$\mathrm{I}={\int }_0^2({\mathrm{f}}^{-1}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)+1)\mathrm{dx}$$$$\mathrm{observe}\mathrm{the}{\int }_0^2{\mathrm{f}}^{-1}\left(\mathrm{x}\right)\mathrm{dx}.\mathrm{Let}{\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\mathrm{u}$$$$\to \mathrm{x}=\mathrm{f}\left(\mathrm{u}\right)\mathrm{and}\mathrm{dx}=\mathrm{f}{}^{\prime }\left(\mathrm{u}\right)\mathrm{so}{\int }_0^2\mathrm{u}\mathrm{f}{}^{\prime }\left(\mathrm{u}\right)\mathrm{du}$$$$=\mathrm{uf}\left(\mathrm{u}\right){\mid }_0^2-{\int }_0^2\mathrm{f}\left(\mathrm{u}\right)\mathrm{du}.$$$$\mathrm{therefore}\mathrm{I}=2\mathrm{f}\left(2\right)+{\int }_0^2\mathrm{dx}=2\mathrm{f}\left(2\right)+2=6$$$$$$

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