Question Number 119214 by abdul88 last updated on 23/Oct/20
$${f}\left({x}\right)\:=\:\left(\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\right){x}^{\mathrm{2}} \:+\:\left(\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}\right){x}\:+\:\left(\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}{f}\left({x}\right){dx}\right)+\mathrm{1} \\ $$$${the}\:{valeu}\:{f}\left(\mathrm{4}\right)\:=\:? \\ $$
Answered by bemath last updated on 23/Oct/20
$${let}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right)\:{dx}\:=\:{p}\:,\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right)\:{dx}\:=\:{q}\:,\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:{f}\left({x}\right)\:{dx}=\:{r} \\ $$$$\Leftrightarrow\:{f}\left({x}\right)\:=\:{px}^{\mathrm{2}} +{qx}+{r}+\mathrm{1} \\ $$$$\Rightarrow\:{p}\:=\frac{\mathrm{1}}{\mathrm{3}}{p}+\frac{\mathrm{1}}{\mathrm{2}}{q}+{r}+\mathrm{1}\:\Rightarrow\mathrm{4}{p}=\mathrm{3}{q}+\mathrm{6}{r}+\mathrm{6} \\ $$$$\Rightarrow{q}\:=\:\frac{\mathrm{8}}{\mathrm{3}}{p}+\mathrm{2}{q}+\mathrm{2}{r}+\mathrm{2}\:\Rightarrow−{q}=\frac{\mathrm{8}}{\mathrm{3}}{p}+\mathrm{2}{r}+\mathrm{2} \\ $$$$\Rightarrow{r}\:=\:\mathrm{9}{p}+\frac{\mathrm{9}}{\mathrm{2}}{q}+\mathrm{3}{r}+\mathrm{3}\:\Rightarrow−\mathrm{2}{r}=\mathrm{9}{p}+\frac{\mathrm{9}}{\mathrm{2}}{q}+\mathrm{3} \\ $$$${we}\:{get}\:\begin{cases}{{p}=\frac{\mathrm{1}}{\mathrm{3}}}\\{{q}=−\frac{\mathrm{80}}{\mathrm{63}}}\\{{r}=−\frac{\mathrm{1}}{\mathrm{7}}}\end{cases} \\ $$$$\because\:{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{80}{x}}{\mathrm{63}}\:+\:\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\Rightarrow{f}\left(\mathrm{4}\right)\:=\:\frac{\mathrm{16}}{\mathrm{3}}−\frac{\mathrm{320}}{\mathrm{63}}+\frac{\mathrm{6}}{\mathrm{7}}=\frac{\mathrm{10}}{\mathrm{9}} \\ $$