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Question Number 157456 by tounghoungko last updated on 23/Oct/21

∫ 5^(3−2x) dx =?

$$\:\int\:\mathrm{5}^{\mathrm{3}−\mathrm{2}{x}} \:{dx}\:=? \\ $$

Answered by FelipeLz last updated on 24/Oct/21

∫5^(3−2x) dx = 5^3 ∫5^(−2x) dx −2x = u dx = −(1/2)du −(5^3 /2)∫5^u du = −(5^3 /2)((5^u /(ln 5)))+c = −(5^(3−2x) /(2ln 5))+c

$$\int\mathrm{5}^{\mathrm{3}−\mathrm{2}{x}} {dx}\:=\:\mathrm{5}^{\mathrm{3}} \int\mathrm{5}^{−\mathrm{2}{x}} {dx} \\ $$$$−\mathrm{2}{x}\:=\:{u} \\ $$$${dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}{du} \\ $$$$−\frac{\mathrm{5}^{\mathrm{3}} }{\mathrm{2}}\int\mathrm{5}^{{u}} {du}\:=\:−\frac{\mathrm{5}^{\mathrm{3}} }{\mathrm{2}}\left(\frac{\mathrm{5}^{{u}} }{\mathrm{ln}\:\mathrm{5}}\right)+{c}\:=\:−\frac{\mathrm{5}^{\mathrm{3}−\mathrm{2}{x}} }{\mathrm{2ln}\:\mathrm{5}}+{c}\: \\ $$

Answered by mahdipoor last updated on 23/Oct/21

=∫e^((3−2x)ln5) dx=∫−(1/(2ln5))e^((3−2x)ln5) (−2ln5)dx= −(1/(2ln5))∫e^((3−2x)ln5) ((3−2x)ln5)′dx= −(1/(2ln5))∫e^u du=−(e^u /(2ln5))=−(e^((3−2x)ln5) /(2ln5))=−(5^(3−2x) /(2ln5))

$$=\int{e}^{\left(\mathrm{3}−\mathrm{2}{x}\right){ln}\mathrm{5}} {dx}=\int−\frac{\mathrm{1}}{\mathrm{2}{ln}\mathrm{5}}{e}^{\left(\mathrm{3}−\mathrm{2}{x}\right){ln}\mathrm{5}} \left(−\mathrm{2}{ln}\mathrm{5}\right){dx}= \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}{ln}\mathrm{5}}\int{e}^{\left(\mathrm{3}−\mathrm{2}{x}\right){ln}\mathrm{5}} \left(\left(\mathrm{3}−\mathrm{2}{x}\right){ln}\mathrm{5}\right)'{dx}= \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}{ln}\mathrm{5}}\int{e}^{{u}} {du}=−\frac{{e}^{{u}} }{\mathrm{2}{ln}\mathrm{5}}=−\frac{{e}^{\left(\mathrm{3}−\mathrm{2}{x}\right){ln}\mathrm{5}} }{\mathrm{2}{ln}\mathrm{5}}=−\frac{\mathrm{5}^{\mathrm{3}−\mathrm{2}{x}} }{\mathrm{2}{ln}\mathrm{5}} \\ $$$$ \\ $$

Commented by tounghoungko last updated on 24/Oct/21

thx

$${thx}\: \\ $$

Answered by mevaa last updated on 23/Oct/21

=∫e^((3−2x)ln 5) dx =∫e^(3ln 5−2xln 5) dx =∫e^(3ln 5) ×e^(−2xln 5) dx =e^(3ln 5) ∫e^(−2xln 5) dx =−e^(3ln 5) /2ln 5∫−2ln 5e^(−2xln 5) dx =−e^(3ln 5) /2ln 5(e^(−2ln 5) )+constante

$$=\int{e}^{\left(\mathrm{3}−\mathrm{2}{x}\right)\mathrm{ln}\:\mathrm{5}} {dx} \\ $$$$=\int{e}^{\mathrm{3ln}\:\mathrm{5}−\mathrm{2}{x}\mathrm{ln}\:\mathrm{5}} {dx} \\ $$$$=\int{e}^{\mathrm{3ln}\:\mathrm{5}} ×{e}^{−\mathrm{2}{x}\mathrm{ln}\:\mathrm{5}} {dx} \\ $$$$={e}^{\mathrm{3ln}\:\mathrm{5}} \int{e}^{−\mathrm{2}{x}\mathrm{ln}\:\mathrm{5}} {dx} \\ $$$$=−{e}^{\mathrm{3ln}\:\mathrm{5}} /\mathrm{2ln}\:\mathrm{5}\int−\mathrm{2ln}\:\mathrm{5}{e}^{−\mathrm{2}{x}\mathrm{ln}\:\mathrm{5}} {dx} \\ $$$$=−{e}^{\mathrm{3ln}\:\mathrm{5}} /\mathrm{2ln}\:\mathrm{5}\left({e}^{−\mathrm{2ln}\:\mathrm{5}} \right)+{constante} \\ $$

Answered by physicstutes last updated on 24/Oct/21

Let 5^(3−2x) = e^y ⇒ y = (3−2x) ln 5 ⇒ dy = −2 ln 5 dx hence dx = −(dy/(2 ln5 )) ⇒ ∫5^(3−2x) dx = −(1/(2 ln 5))∫e^y dy = −(1/(2 ln 5))(5^(3−2x) )+k

$$\mathrm{Let}\:\mathrm{5}^{\mathrm{3}−\mathrm{2}{x}} \:=\:{e}^{{y}} \\ $$$$\Rightarrow\:{y}\:=\:\left(\mathrm{3}−\mathrm{2}{x}\right)\:\mathrm{ln}\:\mathrm{5} \\ $$$$\Rightarrow\:{dy}\:=\:−\mathrm{2}\:\mathrm{ln}\:\mathrm{5}\:{dx} \\ $$$$\:\mathrm{hence}\:{dx}\:=\:−\frac{{dy}}{\mathrm{2}\:\mathrm{ln5}\:} \\ $$$$\Rightarrow\:\int\mathrm{5}^{\mathrm{3}−\mathrm{2}{x}} {dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{ln}\:\mathrm{5}}\int{e}^{{y}} {dy}\:=\:−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{ln}\:\mathrm{5}}\left(\mathrm{5}^{\mathrm{3}−\mathrm{2}{x}} \right)+{k} \\ $$

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