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Question Number 144579 by mohammad17 last updated on 26/Jun/21

Commented by mohammad17 last updated on 26/Jun/21

$$???$$

Answered by liberty last updated on 26/Jun/21

$$\left(1\right)\mathrm{L}={\int }_1^2\sqrt{1+{\left(\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)\right)}^2}\mathrm{dx}$$$$\Rightarrow 8\mathrm{y}{\mathrm{y}}^{\prime }=\frac{32\mathrm{x}{({\mathrm{x}}^2+1)}^2}{3}$$$$\Rightarrow \mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{4\mathrm{x}{({\mathrm{x}}^2+1)}^2}{3\mathrm{y}}$$$$\Rightarrow {\left[\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)\right]}^2=\frac{{16\mathrm{x}}^2{({\mathrm{x}}^2+1)}^4}{4{({\mathrm{x}}^2+1)}^3}={4\mathrm{x}}^2({\mathrm{x}}^2+1)$$$$\mathrm{L}=\underset{1}{\stackrel{2}{\int }}\sqrt{1+{4\mathrm{x}}^4+{4\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{L}=\underset{1}{\stackrel{2}{\int }}\sqrt{{({2\mathrm{x}}^2+1)}^2}\mathrm{dx}$$$$\mathrm{L}=\frac{2}{3}(8-1)+(2-1)=\frac{14}{3}+1=\frac{17}{3}$$

Answered by Dwaipayan Shikari last updated on 26/Jun/21

$$y=\pm \frac{2}{3}{(x^2+1)}^{3/2}$$$$y^{\prime }=\pm 2x{(x^2+1)}^{1/2}$$$${\int }_1^2\sqrt{1+{\left(y^{\prime }\right)}^2}dx$$$$={\int }_1^2\sqrt{1+4x^2(x^2+1)}dx$$$$={\int }_1^2\sqrt{{(2x^2+1)}^2}dx$$$$={\int }_1^2(2x^2+1)dx=\frac{2.7}{3}+1=\frac{17}{3}$$

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