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Question Number 22166 by Joel577 last updated on 12/Oct/17

If∫_1 ^4  f(x) dx = 5  what is the value of ∫_0 ^1  f(3x +1) dx ?

$$\mathrm{If}\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\:{f}\left({x}\right)\:{dx}\:=\:\mathrm{5} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:{f}\left(\mathrm{3}{x}\:+\mathrm{1}\right)\:{dx}\:? \\ $$

Answered by ajfour last updated on 12/Oct/17

I=∫_0 ^(  1) f(3x+1)dx   let 3x+1=t   ⇒  dx=(dt/3)   clearly  then x=0  ⇒ t=1  and when x=1 , t=4  so   I=(1/3)∫_1 ^(  4) f(t)dt =(1/3)∫_1 ^(  4) f(ρ)dρ      =(1/3)∫_1 ^(  4) f(x)dx = (1/3)×5 =(5/3) .

$${I}=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} {f}\left(\mathrm{3}{x}+\mathrm{1}\right){dx}\: \\ $$$${let}\:\mathrm{3}{x}+\mathrm{1}={t}\:\:\:\Rightarrow\:\:{dx}=\frac{{dt}}{\mathrm{3}} \\ $$$$\:{clearly}\:\:{then}\:{x}=\mathrm{0}\:\:\Rightarrow\:{t}=\mathrm{1} \\ $$$${and}\:{when}\:{x}=\mathrm{1}\:,\:{t}=\mathrm{4} \\ $$$${so}\:\:\:{I}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\:\:\mathrm{4}} {f}\left({t}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\:\:\mathrm{4}} {f}\left(\rho\right){d}\rho \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\:\:\mathrm{4}} {f}\left({x}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{5}\:=\frac{\mathrm{5}}{\mathrm{3}}\:. \\ $$

Commented by Joel577 last updated on 13/Oct/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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