Menu Close

Given-f-x-0-1-f-x-dx-x-2-0-2-f-x-dx-x-0-3-f-x-dx-1-then-the-value-of-f-4-




Question Number 123386 by bemath last updated on 25/Nov/20
 Given   f(x)=(∫_0 ^1 f(x)dx)x^2 +(∫_0 ^2 f(x)dx)x+(∫_0 ^3 f(x)dx)+1  then the value of f(4) = ...
$$\:{Given}\: \\ $$$${f}\left({x}\right)=\left(\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\right){x}^{\mathrm{2}} +\left(\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}\right){x}+\left(\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}{f}\left({x}\right){dx}\right)+\mathrm{1} \\ $$$${then}\:{the}\:{value}\:{of}\:{f}\left(\mathrm{4}\right)\:=\:… \\ $$
Answered by TANMAY PANACEA last updated on 25/Nov/20
f(x)=ax^2 +bx+c+1  a=∫_0 ^1 (ax^2 +bx+c+1)  dx  a=∣((ax^3 )/3)+((bx^2 )/2)+cx+x∣_0 ^1   a=(a/3)+(b/2)+c+1.....(1st eqn)  b=∣((ax^3 )/3)+((bx^2 )/2)+cx+x∣_0 ^2   b=((8a)/3)+((4b)/2)+2c+2....+(2nd eqn)  c=∣((ax^3 )/3)+((bx^2 )/2)+cx+x∣_0 ^3   c=((27a)/3)+((9b)/4)+3c+3....(3rd eqn)  we have to solve  to find (a,b and c)  f(x)=ax^2 +bx+c+1  nxt to put x=4    pls try from here..
$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}+\mathrm{1} \\ $$$${a}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({ax}^{\mathrm{2}} +{bx}+{c}+\mathrm{1}\right)\:\:{dx} \\ $$$${a}=\mid\frac{{ax}^{\mathrm{3}} }{\mathrm{3}}+\frac{{bx}^{\mathrm{2}} }{\mathrm{2}}+{cx}+{x}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${a}=\frac{{a}}{\mathrm{3}}+\frac{{b}}{\mathrm{2}}+{c}+\mathrm{1}…..\left(\mathrm{1}{st}\:{eqn}\right) \\ $$$${b}=\mid\frac{{ax}^{\mathrm{3}} }{\mathrm{3}}+\frac{{bx}^{\mathrm{2}} }{\mathrm{2}}+{cx}+{x}\mid_{\mathrm{0}} ^{\mathrm{2}} \\ $$$${b}=\frac{\mathrm{8}{a}}{\mathrm{3}}+\frac{\mathrm{4}{b}}{\mathrm{2}}+\mathrm{2}{c}+\mathrm{2}….+\left(\mathrm{2}{nd}\:{eqn}\right) \\ $$$${c}=\mid\frac{{ax}^{\mathrm{3}} }{\mathrm{3}}+\frac{{bx}^{\mathrm{2}} }{\mathrm{2}}+{cx}+{x}\mid_{\mathrm{0}} ^{\mathrm{3}} \\ $$$${c}=\frac{\mathrm{27}{a}}{\mathrm{3}}+\frac{\mathrm{9}{b}}{\mathrm{4}}+\mathrm{3}{c}+\mathrm{3}….\left(\mathrm{3}{rd}\:{eqn}\right) \\ $$$${we}\:{have}\:{to}\:{solve}\:\:{to}\:{find}\:\left({a},{b}\:{and}\:{c}\right) \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}+\mathrm{1} \\ $$$${nxt}\:{to}\:{put}\:{x}=\mathrm{4} \\ $$$$ \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{try}}\:\boldsymbol{{from}}\:\boldsymbol{{here}}.. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *