A) \[\frac{1}{3}\{4f(1)+f(2)\}\]
B) \[\frac{1}{2}\{f(0)+4f(1)+f(2)\}\]
C) \[\frac{1}{2}\{f(0)+4f(1)\}\]
D) \[\frac{1}{3}\{f(0)+4f(1)+f(2)\}\]
Correct Answer: D
Solution :
[d] Area of \[OABL=\int_{0}^{2}{y\,\,dx}\] \[=\int_{0}^{2}{(a+bx+c{{x}^{2}})dx=2a+2b+\frac{8}{3}c}\] \[=\frac{1}{3}[6a+6b+8c]....(i)\] But, \[f(x)=a+bx+c{{x}^{2}};f(0)=a,f(1)=a+b+c\] \[f(2)=a+2b+4c\Rightarrow \frac{1}{3}\{f(0)+4f(1)+f(2)\}\] \[=\frac{1}{3}\{a+4(a+b+c)+(a+2b+4c)\}\] \[=\frac{1}{3}\{6a+6b+8c\}\]You need to login to perform this action.
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