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Question Number 189066 by cortano12 last updated on 11/Mar/23
Given f(x)+∫_0 ^1 (x+y)^2 f(y) dy=2x^2 −3x+1 find f(x).
$$\:\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} \:\mathrm{f}\left(\mathrm{y}\right)\:\mathrm{dy}=\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{1} \\ $$$$\:\mathrm{find}\:\mathrm{f}\left(\mathrm{x}\right). \\ $$
Answered by horsebrand11 last updated on 11/Mar/23
2x^2 −3x+1=f(x)+∫_0 ^1 (x^2 +2xy+y^2 )dy 2x^2 −3x+1=f(x)+x^2 ∫_0 ^1 f(y)dy+2x∫_0 ^1 y f(y)dy+∫_0 ^1 y^2 f(y) dy let { ((p=∫_0 ^1 f(y)dy)),((q=∫_0 ^1 y f(y)dy )),((r=∫_0 ^1 y^2 f(y)dy )) :} f(x)=2x^2 −px^2 −3x−2qx+1−r f(x)=(2−p)x^2 −(3+2q)x+(1−r)
$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}={f}\left({x}\right)+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){dy} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}={f}\left({x}\right)+{x}^{\mathrm{2}} \underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({y}\right){dy}+\mathrm{2}{x}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{y}\:{f}\left({y}\right){dy}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{y}^{\mathrm{2}} \:{f}\left({y}\right)\:{dy} \\ $$$$\mathrm{let}\:\begin{cases}{{p}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({y}\right){dy}}\\{{q}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{y}\:{f}\left({y}\right){dy}\:}\\{{r}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{y}^{\mathrm{2}} \:{f}\left({y}\right){dy}\:}\end{cases} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −{px}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{2}{qx}+\mathrm{1}−{r} \\ $$$${f}\left({x}\right)=\left(\mathrm{2}−{p}\right){x}^{\mathrm{2}} −\left(\mathrm{3}+\mathrm{2}{q}\right){x}+\left(\mathrm{1}−{r}\right) \\ $$$$ \\ $$
Answered by mr W last updated on 11/Mar/23
f(x)+∫_0 ^1 (x+y)^2 f(y)dy=2x^2 −3x+1 f(x)+∫_0 ^1 (x^2 +2xy+y^2 )f(y)dy=2x^2 −3x+1 f(x)+x^2 ∫_0 ^1 f(y)dy+2x∫_0 ^1 yf(y)dy+∫_0 ^1 y^2 f(y)dy=2x^2 −3x+1 f(x)+Ax^2 +2Bx+C=2x^2 −3x+1 ⇒f(x)=(2−A)x^2 −(3+2B)x+(1−C) A=∫_0 ^1 f(y)dy=∫_0 ^1 [(2−A)y^2 −(3+2B)y+(1−C)]dy A=(2−A)(1/3)−(3+2B)(1/2)+(1−C) ⇒8A+6B+6C=1 ...(i) B=∫_0 ^1 yf(y)dy=∫_0 ^1 [(2−A)y^3 −(3+2B)y^2 +(1−C)y]dy B=(2−A)(1/4)−(3+2B)(1/3)+(1−C)(1/2) ⇒3A+20B+6C=0 ...(ii) C=∫_0 ^1 y^2 f(y)dy=∫_0 ^1 [(2−A)y^4 −(3+2B)y^3 +(1−C)y^2 ]dy C=(2−A)(1/5)−(3+2B)(1/4)+(1−C)(1/3) ⇒12A+30B+80C=−1 ...(iii) from (i),(ii),(iii): A=((71)/(453)),B=−(7/(453)),C=−((356)/(1359)) ⇒f(x)=((835x^2 )/(453))−((1345x)/(453))+((1715)/(1359))
$${f}\left({x}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}+{y}\right)^{\mathrm{2}} {f}\left({y}\right){dy}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1} \\ $$$${f}\left({x}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){f}\left({y}\right){dy}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1} \\ $$$${f}\left({x}\right)+{x}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({y}\right){dy}+\mathrm{2}{x}\int_{\mathrm{0}} ^{\mathrm{1}} {yf}\left({y}\right){dy}+\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}} {f}\left({y}\right){dy}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1} \\ $$$${f}\left({x}\right)+{Ax}^{\mathrm{2}} +\mathrm{2}{Bx}+{C}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)=\left(\mathrm{2}−{A}\right){x}^{\mathrm{2}} −\left(\mathrm{3}+\mathrm{2}{B}\right){x}+\left(\mathrm{1}−{C}\right) \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({y}\right){dy}=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left(\mathrm{2}−{A}\right){y}^{\mathrm{2}} −\left(\mathrm{3}+\mathrm{2}{B}\right){y}+\left(\mathrm{1}−{C}\right)\right]{dy} \\ $$$${A}=\left(\mathrm{2}−{A}\right)\frac{\mathrm{1}}{\mathrm{3}}−\left(\mathrm{3}+\mathrm{2}{B}\right)\frac{\mathrm{1}}{\mathrm{2}}+\left(\mathrm{1}−{C}\right) \\ $$$$\Rightarrow\mathrm{8}{A}+\mathrm{6}{B}+\mathrm{6}{C}=\mathrm{1}\:\:\:…\left({i}\right) \\ $$$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} {yf}\left({y}\right){dy}=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left(\mathrm{2}−{A}\right){y}^{\mathrm{3}} −\left(\mathrm{3}+\mathrm{2}{B}\right){y}^{\mathrm{2}} +\left(\mathrm{1}−{C}\right){y}\right]{dy} \\ $$$${B}=\left(\mathrm{2}−{A}\right)\frac{\mathrm{1}}{\mathrm{4}}−\left(\mathrm{3}+\mathrm{2}{B}\right)\frac{\mathrm{1}}{\mathrm{3}}+\left(\mathrm{1}−{C}\right)\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{A}+\mathrm{20}{B}+\mathrm{6}{C}=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$${C}=\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}} {f}\left({y}\right){dy}=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left(\mathrm{2}−{A}\right){y}^{\mathrm{4}} −\left(\mathrm{3}+\mathrm{2}{B}\right){y}^{\mathrm{3}} +\left(\mathrm{1}−{C}\right){y}^{\mathrm{2}} \right]{dy} \\ $$$${C}=\left(\mathrm{2}−{A}\right)\frac{\mathrm{1}}{\mathrm{5}}−\left(\mathrm{3}+\mathrm{2}{B}\right)\frac{\mathrm{1}}{\mathrm{4}}+\left(\mathrm{1}−{C}\right)\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{12}{A}+\mathrm{30}{B}+\mathrm{80}{C}=−\mathrm{1}\:\:\:…\left({iii}\right) \\ $$$${from}\:\left({i}\right),\left({ii}\right),\left({iii}\right): \\ $$$${A}=\frac{\mathrm{71}}{\mathrm{453}},{B}=−\frac{\mathrm{7}}{\mathrm{453}},{C}=−\frac{\mathrm{356}}{\mathrm{1359}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{835}{x}^{\mathrm{2}} }{\mathrm{453}}−\frac{\mathrm{1345}{x}}{\mathrm{453}}+\frac{\mathrm{1715}}{\mathrm{1359}} \\ $$

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