Question Number 144579 by mohammad17 last updated on 26/Jun/21
Commented by mohammad17 last updated on 26/Jun/21
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Answered by liberty last updated on 26/Jun/21
![(1) L=∫_1 ^( 2) (√(1+(f ′(x))^2 )) dx ⇒8y y′=((32x(x^2 +1)^2 )/3) ⇒f ′(x)= ((4x(x^2 +1)^2 )/(3y)) ⇒[f ′(x)]^2 = ((16x^2 (x^2 +1)^4 )/(4(x^2 +1)^3 ))=4x^2 (x^2 +1) L =∫_1 ^2 (√(1+4x^4 +4x^2 )) dx L = ∫_1 ^2 (√((2x^2 +1)^2 )) dx L= (2/3)(8−1)+(2−1)=((14)/3)+1=((17)/3)](https://www.tinkutara.com/question/Q144585.png)
$$\left(\mathrm{1}\right)\:\mathrm{L}=\int_{\mathrm{1}} ^{\:\mathrm{2}} \sqrt{\mathrm{1}+\left(\mathrm{f}\:'\left(\mathrm{x}\right)\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\:\Rightarrow\mathrm{8y}\:\mathrm{y}'=\frac{\mathrm{32x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{f}\:'\left(\mathrm{x}\right)=\:\frac{\mathrm{4x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3y}} \\ $$$$\Rightarrow\left[\mathrm{f}\:'\left(\mathrm{x}\right)\right]^{\mathrm{2}} =\:\frac{\mathrm{16x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{4x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{L}\:=\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\sqrt{\mathrm{1}+\mathrm{4x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{2}} }\:\mathrm{dx}\: \\ $$$$\mathrm{L}\:=\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\sqrt{\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{L}=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{8}−\mathrm{1}\right)+\left(\mathrm{2}−\mathrm{1}\right)=\frac{\mathrm{14}}{\mathrm{3}}+\mathrm{1}=\frac{\mathrm{17}}{\mathrm{3}} \\ $$
Answered by Dwaipayan Shikari last updated on 26/Jun/21
$${y}=\pm\frac{\mathrm{2}}{\mathrm{3}}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} \\ $$$${y}'=\pm\mathrm{2}{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{\mathrm{1}+\left({y}'\right)^{\mathrm{2}} }\:{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right){dx}=\frac{\mathrm{2}.\mathrm{7}}{\mathrm{3}}+\mathrm{1}=\frac{\mathrm{17}}{\mathrm{3}} \\ $$