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Question Number 95800 by rb222 last updated on 27/May/20

$$iff\left(0\right)=1$$$$f\left(1\right)=2and{\int }_0^{1}f\left(x\right)dx=3$$$$than{\int }_0^{1}xf\left(x\right)dx=?$$$$$$$$a.1$$$$b.-1$$$$c.2$$$$d.-2$$$$$$

Commented by mr W last updated on 27/May/20

$$nouniquesolution!$$

Commented by rb222 last updated on 27/May/20

$$thankssir$$

Answered by bobhans last updated on 27/May/20

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{px}}^2+\mathrm{qx}+1\Rightarrow \mathrm{f}\left(1\right)=\mathrm{p}+\mathrm{q}=1$$$$\mathrm{p}=1-\mathrm{q}\Rightarrow \underset{0}{\stackrel{1}{\int }}((1-\mathrm{q}){\mathrm{x}}^2+\mathrm{qx}+1)\mathrm{dx}=3$$$$\frac{1-\mathrm{q}}{3}+\frac{\mathrm{q}}{2}+1=3;\frac{2+\mathrm{q}}{6}=2\Rightarrow \mathrm{q}=10$$$$\mathrm{p}=-9.\Rightarrow \underset{0}{\stackrel{1}{\int }}\mathrm{xf}\left(\mathrm{x}\right)\mathrm{dx}=\underset{0}{\stackrel{1}{\int }}\mathrm{x}(-{9\mathrm{x}}^2+10\mathrm{x}+1)\mathrm{dx}$$$$=\underset{0}{\stackrel{1}{\int }}(-{9\mathrm{x}}^3+{10\mathrm{x}}^2+\mathrm{x})\mathrm{dx}=\frac{-9}{4}+\frac{10}{3}+\frac{1}{2}$$$$=\frac{-27+40+6}{12}=\frac{19}{12}$$

Commented by mr W last updated on 27/May/20

$$f\left(x\right)canbeanytypeoffunctions.$$

Commented by bobhans last updated on 28/May/20

$$\mathrm{oo}\mathrm{yes}\mathrm{sir}$$

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