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Question Number 79419 by jagoll last updated on 25/Jan/20
given f(x)=f(x+5) ∀x∈R If ∫ _7^9 f(x)=t and ∫ _2^6 f(x)dx = t^2 +4t−3 . find the value of t.
$$\mathrm{given}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}+\mathrm{5}\right)\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{If}\:\int\:_{\mathrm{7}} ^{\mathrm{9}} \:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{t}\:\mathrm{and}\:\int\:_{\mathrm{2}} ^{\mathrm{6}} \:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:= \\ $$$$\mathrm{t}^{\mathrm{2}} +\mathrm{4t}−\mathrm{3}\:.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{t}. \\ $$
Commented by jagoll last updated on 25/Jan/20
∫_7 ^9 f(x−5)d(x−5)=t ∫_2 ^4 f(x)dx= t ⇒ ∫_2 ^6 f(x)dx= ∫_2 ^4 f(x)dx+∫_4 ^6 f(x)dx =2t therefore 2t=t^2 +4t−3 t^2 +2t−3=0 { ((t=−3)),((t=1)) :}
$$\underset{\mathrm{7}} {\overset{\mathrm{9}} {\int}}\mathrm{f}\left(\mathrm{x}−\mathrm{5}\right)\mathrm{d}\left(\mathrm{x}−\mathrm{5}\right)=\mathrm{t} \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\:\mathrm{t}\:\Rightarrow\:\underset{\mathrm{2}} {\overset{\mathrm{6}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}= \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{\mathrm{4}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:=\mathrm{2t} \\ $$$$\mathrm{therefore}\:\mathrm{2t}=\mathrm{t}^{\mathrm{2}} +\mathrm{4t}−\mathrm{3} \\ $$$$\mathrm{t}^{\mathrm{2}} +\mathrm{2t}−\mathrm{3}=\mathrm{0}\:\begin{cases}{\mathrm{t}=−\mathrm{3}}\\{\mathrm{t}=\mathrm{1}}\end{cases} \\ $$
Commented by jagoll last updated on 25/Jan/20
Mr W , that right?
$$\mathrm{Mr}\:\mathrm{W}\:,\:\mathrm{that}\:\mathrm{right}? \\ $$
Commented by john santu last updated on 25/Jan/20
i think wrong. ∫_2 ^6 f(x)dx=∫_2 ^6 f(x+5)d(x+5)= ∫_7 ^(11) f(x)dx = ??
$${i}\:{think}\:{wrong}. \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{6}} {\int}}{f}\left({x}\right){dx}=\underset{\mathrm{2}} {\overset{\mathrm{6}} {\int}}{f}\left({x}+\mathrm{5}\right){d}\left({x}+\mathrm{5}\right)= \\ $$$$\underset{\mathrm{7}} {\overset{\mathrm{11}} {\int}}{f}\left({x}\right){dx}\:=\:?? \\ $$
Commented by mr W last updated on 25/Jan/20
it′s wrong sir! ∫_2 ^6 f(x)dx=∫_2 ^4 f(x)dx+∫_4 ^6 f(x)dx =t+∫_4 ^6 f(x)dx ≠2t since ∫_4 ^6 f(x)dx≠∫_2 ^4 f(x)dx
$${it}'{s}\:{wrong}\:{sir}! \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{6}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{\mathrm{4}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\: \\ $$$$={t}+\underset{\mathrm{4}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:\neq\mathrm{2t} \\ $$$${since}\:\underset{\mathrm{4}} {\overset{\mathrm{6}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\neq\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$
Commented by mr W last updated on 25/Jan/20
again, if T is period, ∫_a ^(a+nT) f(x)dx=∫_c ^(c+nT) f(x)dx for any a and c but ∫_a ^(a+S) f(x)dx≠∫_c ^(c+S) f(x)dx if S≠nT and c≠a+kT
$${again},\:{if}\:{T}\:{is}\:{period}, \\ $$$$\int_{{a}} ^{{a}+{nT}} {f}\left({x}\right){dx}=\int_{{c}} ^{{c}+{nT}} {f}\left({x}\right){dx}\:{for}\:{any}\:{a}\:{and}\:{c} \\ $$$${but}\:\int_{{a}} ^{{a}+{S}} {f}\left({x}\right){dx}\neq\int_{{c}} ^{{c}+{S}} {f}\left({x}\right){dx}\:{if}\:{S}\neq{nT}\:{and}\:{c}\neq{a}+{kT} \\ $$
Commented by jagoll last updated on 25/Jan/20
thanks you mister
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{mister} \\ $$

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