Question Number 124853 by bemath last updated on 06/Dec/20
$$\:\:\int_{\mathrm{1}} ^{\:{x}^{\mathrm{3}} +\mathrm{5}{x}} {f}\left({t}\right)\:{dt}\:=\:\mathrm{2}{x}\:\: \\ $$$$\:{then}\:{f}\left(\mathrm{18}\right)\:=? \\ $$
Answered by liberty last updated on 06/Dec/20
![(d/dx) [ ∫_( 1) ^( x^3 +5x) f(t) dt ] = (d/dx)(2x) (3x^2 +5) f(x^3 +5x) = 2 ⇒ f(x^3 +5x) = (2/(3x^2 +5)) put x^3 +5x = 18 ; (x−2)(x^2 +2x+9)=0 ⇒x_1 = 2 ∧ f(18) = (2/(3×4+5)) = (2/(17)) ⇒x_(2,3) = ((−2±4i(√2))/2) = −1±2i(√2) we get f(18) = (2/(3(−7−4i(√2))+5))=(2/(−16−12i(√2))) =(1/(−8−6i(√2))) and f(18) = (2/(3(−7+4i(√2))+5)) = (2/(−16+12i(√2))) = (1/(−8+6i(√2)))](Q124854.png)
$$\:\:\frac{{d}}{{dx}}\:\left[\:\int_{\:\mathrm{1}} ^{\:{x}^{\mathrm{3}} +\mathrm{5}{x}} {f}\left({t}\right)\:{dt}\:\right]\:=\:\frac{{d}}{{dx}}\left(\mathrm{2}{x}\right) \\ $$$$\:\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}\right)\:{f}\left({x}^{\mathrm{3}} +\mathrm{5}{x}\right)\:=\:\mathrm{2}\:\Rightarrow\:{f}\left({x}^{\mathrm{3}} +\mathrm{5}{x}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}} \\ $$$$\:{put}\:{x}^{\mathrm{3}} +\mathrm{5}{x}\:=\:\mathrm{18}\:;\:\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{9}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1}} \:=\:\mathrm{2}\:\wedge\:{f}\left(\mathrm{18}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}×\mathrm{4}+\mathrm{5}}\:=\:\frac{\mathrm{2}}{\mathrm{17}} \\ $$$$\Rightarrow{x}_{\mathrm{2},\mathrm{3}} \:=\:\frac{−\mathrm{2}\pm\mathrm{4}{i}\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:−\mathrm{1}\pm\mathrm{2}{i}\sqrt{\mathrm{2}} \\ $$$${we}\:{get}\:{f}\left(\mathrm{18}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}\left(−\mathrm{7}−\mathrm{4}{i}\sqrt{\mathrm{2}}\right)+\mathrm{5}}=\frac{\mathrm{2}}{−\mathrm{16}−\mathrm{12}{i}\sqrt{\mathrm{2}}}\:=\frac{\mathrm{1}}{−\mathrm{8}−\mathrm{6}{i}\sqrt{\mathrm{2}}} \\ $$$${and}\:{f}\left(\mathrm{18}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}\left(−\mathrm{7}+\mathrm{4}{i}\sqrt{\mathrm{2}}\right)+\mathrm{5}}\:=\:\frac{\mathrm{2}}{−\mathrm{16}+\mathrm{12}{i}\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{−\mathrm{8}+\mathrm{6}{i}\sqrt{\mathrm{2}}} \\ $$
Answered by mathmax by abdo last updated on 06/Dec/20
$$\mathrm{by}\:\mathrm{derivation}\:\mathrm{we}\:\mathrm{get}\:\:\left(\mathrm{3x}^{\mathrm{2}} \:+\mathrm{5}\right)\mathrm{f}\left(\mathrm{x}^{\mathrm{3}} \:+\mathrm{5x}\right)=\mathrm{2} \\ $$$$\mathrm{x}=\mathrm{2}\:\Rightarrow\left(\mathrm{12}+\mathrm{5}\right)\mathrm{f}\left(\mathrm{8}+\mathrm{10}\right)=\mathrm{2}\:\Rightarrow\mathrm{17f}\left(\mathrm{18}\right)=\mathrm{2}\:\Rightarrow\mathrm{f}\left(\mathrm{18}\right)=\frac{\mathrm{2}}{\mathrm{17}} \\ $$