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Question Number 32323 by Joel578 last updated on 23/Mar/18

$$\mathrm{Given}$$$$f\left(x\right)=\frac{3}{16}\left({\int }_0^{1}f\left(x\right)dx\right)x^2-\frac{9}{10}\left({\int }_0^{2}f\left(x\right)dx\right)x+2\left({\int }_0^{3}f\left(x\right)dx\right)+4$$$$\mathrm{Find}f\left(x\right)$$

Answered by ajfour last updated on 23/Mar/18

$$letf\left(x\right)={\mathit{A}\mathit{x}}^2+\mathit{B}x+\mathit{C}$$$$\Rightarrow {Ax}^2+Bx+C=\frac{3}{16}(\frac{A}{3}+\frac{B}{2}+C)x^2$$$$-\frac{9}{10}(\frac{8A}{3}+2B+2C)x$$$$+2(\frac{27A}{3}+\frac{9B}{2}+3C)+4$$$$forx=0$$$$C=18A+9B+6C+4$$$$\Rightarrow 18A+9B+5C=-4$$$$forx=1$$$$A+B+C=(\frac{1}{16}-\frac{12}{5}+18)A+$$$$(\frac{3}{32}-\frac{9}{5}+9)B+(\frac{3}{16}-\frac{9}{5}+6)C+4$$$$\Rightarrow 160(A+B+C)=......$$

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