question_answer

If \[f(x)=a+bx+c{{x}^{2}}\], where \[c>0\] and \[{{b}^{2}}-4ac<0\], then the area enclosed by the coordinate axes, the line \[x=2\] and the curve \[y=f(x)\] is given by

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, \[f(x)=a+bx+c{{x}^{2}};f(0)=a,f(1)=a+b+c\] \[f(2)=a+2b+4c\Rightarrow \frac{1}{3}\{f(0)+4f(1)+f(2)\}\] \[=\frac{1}{3}\{a+4(a+b+c)+(a+2b+4c)\}\] \[=\frac{1}{3}\{6a+6b+8c\}\]