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Question Number 163891 by HongKing last updated on 11/Jan/22

$$\mathrm{if}\mathrm{f}({\mathrm{x}}^3+1)={\mathrm{x}}^5+4\mathrm{x}+2$$$$\mathrm{find}\underset{0}{\stackrel{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=?$$

Answered by Ar Brandon last updated on 11/Jan/22

$$x^3+1=y\Rightarrow x=\sqrt[3]{y-1}$$$$\Rightarrow f\left(y\right)={(y-1)}^{\frac{5}{3}}+4{(y-1)}^{\frac{1}{3}}+2$$$$\Rightarrow f\left(x\right)={(x-1)}^{\frac{5}{3}}+4{(x-1)}^{\frac{1}{3}}+2$$$${\int }_0^{1}f\left(x\right)dx={[\frac{3}{8}{(x-1)}^{\frac{8}{3}}+3{(x-1)}^{\frac{4}{3}}+2x]}_0^1$$$$=2-(\frac{3}{8}+3)=-\frac{11}{8}$$

Commented by HongKing last updated on 11/Jan/22

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Answered by mr W last updated on 11/Jan/22

$${\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$$$$={\int }_{-1}^0\mathrm{f}({\mathrm{t}}^3+1)\mathrm{d}({\mathrm{t}}^3+1)$$$$={\int }_{-1}^0{3\mathrm{t}}^2({\mathrm{t}}^5+4\mathrm{t}+2)\mathrm{dt}$$$$=3{[\frac{{\mathrm{t}}^8}{8}+{\mathrm{t}}^4+\frac{{2\mathrm{t}}^3}{3}]}_{-1}^0$$$$=3(-\frac{1}{8}-1+\frac{2}{3})$$$$=-\frac{11}{8}$$

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