if-f-0-1-f-1-2-and-0-1-f-x-dx-3-than-0-1-x-f-x-dx-a-1-b-1-c-2-d-2-

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

$${if}\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2}\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right)\:{dx}=\mathrm{3}\: \\ $$$${than}\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}\:{f}\left({x}\right)\:{dx}\:=\:? \\ $$$$ \\ $$$${a}.\:\mathrm{1} \\ $$$${b}.\:−\mathrm{1} \\ $$$${c}.\:\mathrm{2} \\ $$$${d}.\:−\mathrm{2} \\ $$$$ \\ $$

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

$${no}\:{unique}\:{solution}! \\ $$

Commented by rb222 last updated on 27/May/20

$${thanks}\:{sir} \\ $$

Answered by bobhans last updated on 27/May/20

f(x) = px^2 +qx+1 ⇒f(1) = p+q = 1 p= 1−q ⇒ ∫_0 ^1 ((1−q)x^2 +qx+1)dx = 3 ((1−q)/3)+(q/2)+1 = 3 ; ((2+q)/6) = 2 ⇒q=10 p=−9. ⇒∫_0 ^1 xf(x)dx = ∫_0 ^1 x(−9x^2 +10x+1)dx =∫_0 ^1 (−9x^3 +10x^2 +x) dx = ((−9)/4)+((10)/3)+(1/2) = ((−27+40+6)/(12)) = ((19)/(12))

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

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

$${f}\left({x}\right)\:{can}\:{be}\:{any}\:{type}\:{of}\:{functions}. \\ $$

Commented by bobhans last updated on 28/May/20

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

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