Question Number 145525 by mathdanisur last updated on 05/Jul/21
Answered by Olaf_Thorendsen last updated on 05/Jul/21
![f(x+2)+10f(x) = 7f(x+1) f(x+2)−7f(x+1)+10f(x) = 0 r^2 −7r+10 = 0 (r−2)(r−5) = 0 f(x) = λ.2^x +μ5^x f(0) = λ+μ = 2 f(1) = 2λ+5μ = 7 λ = 1, μ = 1 f(x) = 2^x +5^x f(x) = e^(xln2) +e^(xln5) ∫_0 ^1 f(x)dx = [(2^x /(ln2))+(5^x /(ln5))]_0 ^1 = (1/(ln2))+(4/(ln5)) ∫_0 ^1 f(x)dx = ((ln5+4ln2)/(ln2.ln5)) = ((ln80)/(ln2.ln5)) ln2.ln5∫_0 ^1 f(x)dx = ln80](https://www.tinkutara.com/question/Q145532.png)
$${f}\left({x}+\mathrm{2}\right)+\mathrm{10}{f}\left({x}\right)\:=\:\mathrm{7}{f}\left({x}+\mathrm{1}\right) \\ $$$${f}\left({x}+\mathrm{2}\right)−\mathrm{7}{f}\left({x}+\mathrm{1}\right)+\mathrm{10}{f}\left({x}\right)\:=\:\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{7}{r}+\mathrm{10}\:=\:\mathrm{0} \\ $$$$\left({r}−\mathrm{2}\right)\left({r}−\mathrm{5}\right)\:=\:\mathrm{0} \\ $$$${f}\left({x}\right)\:=\:\lambda.\mathrm{2}^{{x}} +\mu\mathrm{5}^{{x}} \\ $$$${f}\left(\mathrm{0}\right)\:=\:\lambda+\mu\:=\:\mathrm{2} \\ $$$${f}\left(\mathrm{1}\right)\:=\:\mathrm{2}\lambda+\mathrm{5}\mu\:=\:\mathrm{7} \\ $$$$\lambda\:=\:\mathrm{1},\:\mu\:=\:\mathrm{1} \\ $$$${f}\left({x}\right)\:=\:\mathrm{2}^{{x}} +\mathrm{5}^{{x}} \\ $$$${f}\left({x}\right)\:=\:{e}^{{x}\mathrm{ln2}} +{e}^{{x}\mathrm{ln5}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:=\:\left[\frac{\mathrm{2}^{{x}} }{\mathrm{ln2}}+\frac{\mathrm{5}^{{x}} }{\mathrm{ln5}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{ln2}}+\frac{\mathrm{4}}{\mathrm{ln5}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:=\:\frac{\mathrm{ln5}+\mathrm{4ln2}}{\mathrm{ln2}.\mathrm{ln5}}\:=\:\frac{\mathrm{ln80}}{\mathrm{ln2}.\mathrm{ln5}} \\ $$$$\mathrm{ln2}.\mathrm{ln5}\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:=\:\mathrm{ln80} \\ $$
Commented by mathdanisur last updated on 05/Jul/21
$${cool}\:{alot}\:{Ser}\:{thankyou} \\ $$