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Question Number 78717 by jagoll last updated on 20/Jan/20

given   ∫ f(x) dx = (1/(2 ((g(x)))^(1/(3 )) )) .   g′(1)= g(1) = 8 ⇒f(1)=?

$$\mathrm{given}\: \\ $$$$\int\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}\:\sqrt[{\mathrm{3}\:}]{\mathrm{g}\left(\mathrm{x}\right)}}\:.\: \\ $$$$\mathrm{g}'\left(\mathrm{1}\right)=\:\mathrm{g}\left(\mathrm{1}\right)\:=\:\mathrm{8}\:\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=? \\ $$$$ \\ $$

Commented by john santu last updated on 20/Jan/20

∫f(x)dx = (1/2)(g(x))^(−(1/3))   f(x)= −(1/6)g′(x)×(1/(((g(x))^4 ))^(1/(3 )) )  f(1)= −(1/6).(8).(1/(16))=−(1/(12)) ★

$$\int{f}\left({x}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({g}\left({x}\right)\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${f}\left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{6}}{g}'\left({x}\right)×\frac{\mathrm{1}}{\sqrt[{\mathrm{3}\:}]{\left({g}\left({x}\right)\right)^{\mathrm{4}} }} \\ $$$${f}\left(\mathrm{1}\right)=\:−\frac{\mathrm{1}}{\mathrm{6}}.\left(\mathrm{8}\right).\frac{\mathrm{1}}{\mathrm{16}}=−\frac{\mathrm{1}}{\mathrm{12}}\:\bigstar \\ $$

Commented by jagoll last updated on 20/Jan/20

thanks

$$\mathrm{thanks} \\ $$

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