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Question Number 119214 by abdul88 last updated on 23/Oct/20

f(x) = (∫_0 ^1 f(x)dx)x^2 + (∫_0 ^2 f(x)dx)x + (∫_0 ^3 f(x)dx)+1 the valeu f(4) = ?

$${f}\left({x}\right)\:=\:\left(\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\right){x}^{\mathrm{2}} \:+\:\left(\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}\right){x}\:+\:\left(\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}{f}\left({x}\right){dx}\right)+\mathrm{1} \\ $$$${the}\:{valeu}\:{f}\left(\mathrm{4}\right)\:=\:? \\ $$

Answered by bemath last updated on 23/Oct/20

let ∫_0 ^1 f(x) dx = p , ∫_0 ^2 f(x) dx = q , ∫_0 ^3 f(x) dx= r ⇔ f(x) = px^2 +qx+r+1 ⇒ p =(1/3)p+(1/2)q+r+1 ⇒4p=3q+6r+6 ⇒q = (8/3)p+2q+2r+2 ⇒−q=(8/3)p+2r+2 ⇒r = 9p+(9/2)q+3r+3 ⇒−2r=9p+(9/2)q+3 we get { ((p=(1/3))),((q=−((80)/(63)))),((r=−(1/7))) :} ∵ f(x) = (x^2 /3)−((80x)/(63)) + (6/7) ⇒f(4) = ((16)/3)−((320)/(63))+(6/7)=((10)/9)

$${let}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right)\:{dx}\:=\:{p}\:,\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right)\:{dx}\:=\:{q}\:,\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:{f}\left({x}\right)\:{dx}=\:{r} \\ $$$$\Leftrightarrow\:{f}\left({x}\right)\:=\:{px}^{\mathrm{2}} +{qx}+{r}+\mathrm{1} \\ $$$$\Rightarrow\:{p}\:=\frac{\mathrm{1}}{\mathrm{3}}{p}+\frac{\mathrm{1}}{\mathrm{2}}{q}+{r}+\mathrm{1}\:\Rightarrow\mathrm{4}{p}=\mathrm{3}{q}+\mathrm{6}{r}+\mathrm{6} \\ $$$$\Rightarrow{q}\:=\:\frac{\mathrm{8}}{\mathrm{3}}{p}+\mathrm{2}{q}+\mathrm{2}{r}+\mathrm{2}\:\Rightarrow−{q}=\frac{\mathrm{8}}{\mathrm{3}}{p}+\mathrm{2}{r}+\mathrm{2} \\ $$$$\Rightarrow{r}\:=\:\mathrm{9}{p}+\frac{\mathrm{9}}{\mathrm{2}}{q}+\mathrm{3}{r}+\mathrm{3}\:\Rightarrow−\mathrm{2}{r}=\mathrm{9}{p}+\frac{\mathrm{9}}{\mathrm{2}}{q}+\mathrm{3} \\ $$$${we}\:{get}\:\begin{cases}{{p}=\frac{\mathrm{1}}{\mathrm{3}}}\\{{q}=−\frac{\mathrm{80}}{\mathrm{63}}}\\{{r}=−\frac{\mathrm{1}}{\mathrm{7}}}\end{cases} \\ $$$$\because\:{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{80}{x}}{\mathrm{63}}\:+\:\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\Rightarrow{f}\left(\mathrm{4}\right)\:=\:\frac{\mathrm{16}}{\mathrm{3}}−\frac{\mathrm{320}}{\mathrm{63}}+\frac{\mathrm{6}}{\mathrm{7}}=\frac{\mathrm{10}}{\mathrm{9}} \\ $$

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