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Question Number 134391 by bramlexs22 last updated on 03/Mar/21

$$\mathrm{Given}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^x\left(\frac{t^4-t^2}{t^2+1}\right)dt.$$$$\mathrm{Find}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right).$$

Answered by liberty last updated on 03/Mar/21

$$\mathrm{Given}\mathrm{f}\left(\mathrm{x}\right)={\int }_0^x\left[\frac{t^4-t^2}{t^2+1}\right]dt.$$$$\mathrm{Find}\mathrm{minimum}\mathrm{value}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right).$$$$(\bullet )\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\frac{{\mathrm{x}}^4-{\mathrm{x}}^2}{{\mathrm{x}}^2+1}=0$$$${\mathrm{x}}^2(\mathrm{x}-1)=0\to \{\begin{array}{l}\mathrm{x}=0\\ \mathrm{x}=1\end{array}$$$$(\bullet \bullet )\frac{{\mathrm{d}}^2\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}{\mid }_{\mathrm{x}=1}=\frac{({4\mathrm{x}}^3-2\mathrm{x})({\mathrm{x}}^2+1)-2\mathrm{x}({\mathrm{x}}^4-{\mathrm{x}}^2)}{{({\mathrm{x}}^2+1)}^2}>0\mathrm{for}\mathrm{x}=1$$$$\mathrm{so}\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{f}\left(1\right)$$$$(\bullet \bullet \bullet )\mathrm{f}\left(1\right)={\int }_0^1\frac{{\mathrm{t}}^4-{\mathrm{t}}^2}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$\mathrm{f}\left(1\right)={\int }_0^1({\mathrm{t}}^2-2+\frac{2}{{\mathrm{t}}^2+1})\mathrm{dt}$$$$\mathrm{f}\left(1\right)={[\frac{{\mathrm{t}}^3}{3}-2\mathrm{t}+2\arctan \mathrm{t}]}_0^1$$$$\mathrm{f}\left(1\right)=\frac{1}{3}-2+2\left(\frac{\pi }{4}\right)=\frac{\pi }{2}-\frac{5}{6}$$$$$$

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