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Question Number 143588 by bobhans last updated on 16/Jun/21

Answered by EDWIN88 last updated on 16/Jun/21

$$\mathrm{F}\left(\mathrm{x}\right)=\underset{4}{\stackrel{8\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\sqrt{2+{\mathrm{x}}^2}+\mathrm{c}$$$$\mathrm{F}{}^{\prime }\left(\mathrm{x}\right)=8\mathrm{f}\left(8\mathrm{x}\right)=\frac{\mathrm{x}}{\sqrt{2+{\mathrm{x}}^2}}$$$${\mathrm{F}}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(8\mathrm{x}\right)=\frac{\mathrm{x}}{8\sqrt{2+{\mathrm{x}}^2}}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\frac{1}{8}\mathrm{x}}{8\sqrt{2+\frac{1}{64}{\mathrm{x}}^2}}=\frac{\mathrm{x}}{8\sqrt{128+{\mathrm{x}}^2}}$$$$\Rightarrow \mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{8\sqrt{128+{\mathrm{t}}^2}}$$$$\mathrm{F}\left(\mathrm{x}\right)=\underset{4}{\stackrel{8\mathrm{x}}{\int }}\left[\frac{\mathrm{t}}{8\sqrt{128+{\mathrm{t}}^2}}\right]\mathrm{dt}=\sqrt{2+{\mathrm{x}}^2}+\mathrm{c}$$$$\mathrm{take}\mathrm{x}=\frac{1}{2}\Rightarrow \mathrm{F}\left(\frac{1}{2}\right)=\underset{4}{\stackrel{8.\frac{1}{2}}{\int }}\left[\frac{\mathrm{t}}{8\sqrt{128+{\mathrm{t}}^2}}\right]\mathrm{dt}=\sqrt{2+\frac{1}{4}}+\mathrm{c}$$$$\Rightarrow 0=\frac{3}{2}+\mathrm{c};\mathrm{c}=-\frac{3}{2}$$

Answered by mathmax by abdo last updated on 16/Jun/21

$${\int }_4^{8\mathrm{x}}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\sqrt{2+{\mathrm{x}}^2}+\mathrm{c}\mathrm{by}\mathrm{derivation}\mathrm{we}\mathrm{get}$$$$8\mathrm{f}\left(8\mathrm{x}\right)=\frac{2\mathrm{x}}{2\sqrt{2+{\mathrm{x}}^2}}=\frac{\mathrm{x}}{\sqrt{2+{\mathrm{x}}^2}}\Rightarrow \mathrm{f}\left(8\mathrm{x}\right)=\frac{\mathrm{x}}{8\sqrt{2+{\mathrm{x}}^2}}$$$$8\mathrm{x}=\mathrm{t}\Rightarrow \mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{64\sqrt{2+\frac{{\mathrm{t}}^2}{8^2}}}=\frac{\mathrm{t}}{8\sqrt{128+{\mathrm{t}}^2}}$$$$\mathrm{x}=0\Rightarrow -{\int }_0^4\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\sqrt{2}+\mathrm{c}\Rightarrow \mathrm{c}=-\sqrt{2}-{\int }_0^4\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$$$$=-\sqrt{2}-\frac{1}{8}{\int }_0^4\frac{\mathrm{t}}{\sqrt{128+{\mathrm{t}}^2}}\mathrm{dt}=-\sqrt{2}{\left[\sqrt{128+{\mathrm{t}}^2}\right]}_0^4$$$$=-\sqrt{2}(\sqrt{128+16}-\sqrt{128})=-\sqrt{2}(\sqrt{144}-\sqrt{128})$$

Answered by mindispower last updated on 16/Jun/21

$$x=\frac{1}{2}use{\int }_s^{s}f\left(x\right)dx=0$$

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