A) \[2\left\{ 3f(1)+2f\left( \frac{1}{2} \right) \right\}\]
B) \[\frac{1}{2}\left\{ f(1)+3f\left( \frac{1}{2} \right) \right\}\]
C) \[\frac{1}{6}\left\{ f(0)+f(1)+4f\left( \frac{1}{2} \right) \right\}\]
D) \[\frac{1}{3}\left\{ f(0)+f\left( \frac{1}{2} \right) \right\}\]
Correct Answer: C
Solution :
[c] \[\int_{0}^{1}{(a+bx+c{{x}^{2}})dx=\left[ ax+\frac{b{{x}^{2}}}{2}+\frac{c{{x}^{3}}}{3} \right]_{0}^{1}}\] \[=\frac{1}{6}(6a+3b+2c)\] \[\because \,\,\,\,\,\,\,\,\,\,f(0)=a\] ??(i) \[f(1)=a+b+c\] ?..(ii) \[f\left( \frac{1}{2} \right)=a+\frac{b}{2}+\frac{c}{4}\] \[\Rightarrow \,\,\,\,\,\,4f\left( \frac{1}{2} \right)=4a+2b+c\] ??(iii) \[(i)+(ii)+(iii)\] \[6a+3b+2c=f(0)+f(1)+4f\left( \frac{1}{2} \right)\] Hence, \[\int_{0}^{1}{f(x)\,dx=\frac{1}{6}}\left( f(0)+f(1)+4f\left( \frac{1}{2} \right) \right)\]
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