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

Given f(x) = (3/(16))(∫_0 ^1 f(x)dx)x^2 − (9/(10))(∫_0 ^2 f(x)dx)x + 2(∫_0 ^3 f(x)dx) + 4 Find f(x)

$$\mathrm{Given} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{16}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\right){x}^{\mathrm{2}} \:−\:\frac{\mathrm{9}}{\mathrm{10}}\left(\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right){dx}\right){x}\:+\:\mathrm{2}\left(\int_{\mathrm{0}} ^{\mathrm{3}} {f}\left({x}\right){dx}\right)\:+\:\mathrm{4} \\ $$$$\mathrm{Find}\:{f}\left({x}\right) \\ $$

Answered by ajfour last updated on 23/Mar/18

let f(x)=Ax^2 +Bx+C ⇒ Ax^2 +Bx+C=(3/(16))((A/3)+(B/2)+C)x^2 −(9/(10))(((8A)/3)+2B+2C)x +2(((27A)/3)+((9B)/2)+3C)+4 for x=0 C=18A+9B+6C+4 ⇒ 18A+9B+5C=−4 for x=1 A+B+C=((1/(16))−((12)/5)+18)A+ ((3/(32))−(9/5)+9)B+((3/(16))−(9/5)+6)C+4 ⇒ 160(A+B+C)=......

$${let}\:\:{f}\left({x}\right)=\boldsymbol{{Ax}}^{\mathrm{2}} +\boldsymbol{{B}}{x}+\boldsymbol{{C}} \\ $$$$\Rightarrow\:{Ax}^{\mathrm{2}} +{Bx}+{C}=\frac{\mathrm{3}}{\mathrm{16}}\left(\frac{{A}}{\mathrm{3}}+\frac{{B}}{\mathrm{2}}+{C}\right){x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{9}}{\mathrm{10}}\left(\frac{\mathrm{8}{A}}{\mathrm{3}}+\mathrm{2}{B}+\mathrm{2}{C}\right){x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\left(\frac{\mathrm{27}{A}}{\mathrm{3}}+\frac{\mathrm{9}{B}}{\mathrm{2}}+\mathrm{3}{C}\right)+\mathrm{4} \\ $$$${for}\:\:{x}=\mathrm{0} \\ $$$${C}=\mathrm{18}{A}+\mathrm{9}{B}+\mathrm{6}{C}+\mathrm{4} \\ $$$$\Rightarrow\:\:\:\:\mathrm{18}{A}+\mathrm{9}{B}+\mathrm{5}{C}=−\mathrm{4} \\ $$$${for}\:\:\:{x}=\mathrm{1} \\ $$$${A}+{B}+{C}=\left(\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{12}}{\mathrm{5}}+\mathrm{18}\right){A}+ \\ $$$$\:\:\:\:\:\:\:\:\left(\frac{\mathrm{3}}{\mathrm{32}}−\frac{\mathrm{9}}{\mathrm{5}}+\mathrm{9}\right){B}+\left(\frac{\mathrm{3}}{\mathrm{16}}−\frac{\mathrm{9}}{\mathrm{5}}+\mathrm{6}\right){C}+\mathrm{4} \\ $$$$\Rightarrow\:\:\mathrm{160}\left({A}+{B}+{C}\right)=...... \\ $$

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