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Question Number 189066 by cortano12 last updated on 11/Mar/23

$$\mathrm{Given}\mathrm{f}\left(\mathrm{x}\right)+\underset{0}{\stackrel{1}{\int }}{(\mathrm{x}+\mathrm{y})}^2\mathrm{f}\left(\mathrm{y}\right)\mathrm{dy}={2\mathrm{x}}^2-3\mathrm{x}+1$$$$\mathrm{find}\mathrm{f}\left(\mathrm{x}\right).$$

Answered by horsebrand11 last updated on 11/Mar/23

$$2x^2-3x+1=f\left(x\right)+\underset{0}{\stackrel{1}{\int }}(x^2+2xy+y^2)dy$$$$2x^2-3x+1=f\left(x\right)+x^2\underset{0}{\stackrel{1}{\int }}f\left(y\right)dy+2x\underset{0}{\stackrel{1}{\int }}yf\left(y\right)dy+\underset{0}{\stackrel{1}{\int }}{y}^{2}f\left(y\right)dy$$$$\mathrm{let}\{\begin{array}{l}p=\underset{0}{\stackrel{1}{\int }}f\left(y\right)dy\\ q=\underset{0}{\stackrel{1}{\int }}yf\left(y\right)dy\\ r=\underset{0}{\stackrel{1}{\int }}{y}^{2}f\left(y\right)dy\end{array}$$$$f\left(x\right)=2x^2-{px}^2-3x-2qx+1-r$$$$f\left(x\right)=(2-p)x^2-(3+2q)x+(1-r)$$$$$$

Answered by mr W last updated on 11/Mar/23

$$f\left(x\right)+{\int }_0^1{(x+y)}^{2}f\left(y\right)dy=2x^2-3x+1$$$$f\left(x\right)+{\int }_0^1(x^2+2xy+y^2)f\left(y\right)dy=2x^2-3x+1$$$$f\left(x\right)+x^2{\int }_0^{1}f\left(y\right)dy+2x{\int }_0^{1}yf\left(y\right)dy+{\int }_0^1{y}^{2}f\left(y\right)dy=2x^2-3x+1$$$$f\left(x\right)+{Ax}^2+2Bx+C=2x^2-3x+1$$$$\Rightarrow f\left(x\right)=(2-A)x^2-(3+2B)x+(1-C)$$$$A={\int }_0^{1}f\left(y\right)dy={\int }_0^1[(2-A)y^2-(3+2B)y+(1-C)]dy$$$$A=(2-A)\frac{1}{3}-(3+2B)\frac{1}{2}+(1-C)$$$$\Rightarrow 8A+6B+6C=1...\left(i\right)$$$$B={\int }_0^{1}yf\left(y\right)dy={\int }_0^1[(2-A)y^3-(3+2B)y^2+(1-C)y]dy$$$$B=(2-A)\frac{1}{4}-(3+2B)\frac{1}{3}+(1-C)\frac{1}{2}$$$$\Rightarrow 3A+20B+6C=0...\left(ii\right)$$$$C={\int }_0^1{y}^{2}f\left(y\right)dy={\int }_0^1[(2-A)y^4-(3+2B)y^3+(1-C)y^2]dy$$$$C=(2-A)\frac{1}{5}-(3+2B)\frac{1}{4}+(1-C)\frac{1}{3}$$$$\Rightarrow 12A+30B+80C=-1...\left(iii\right)$$$$from\left(i\right),\left(ii\right),\left(iii\right):$$$$A=\frac{71}{453},B=-\frac{7}{453},C=-\frac{356}{1359}$$$$\Rightarrow f\left(x\right)=\frac{835x^2}{453}-\frac{1345x}{453}+\frac{1715}{1359}$$

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