A) \[[f(0)+4f(1/2)+f(1)]/6\]
B) \[[f(0)+4f(1/2)+f(1)]/3\]
C) \[[f(0)+4f(1/2)+f(1)]\]
D) \[[f(0)+2f(1/2)+f(1)]/6\]
Correct Answer: A
Solution :
| [a] Given, \[f(x)=a+bx+c{{x}^{2}}\] |
| \[\therefore \,\,\,\,\int_{0}^{1}{f(x)dx=\int_{0}^{1}{(a+bx+c{{x}^{2}})dx}}\] |
| \[=\left[ ax+\frac{b{{x}^{2}}}{2}+\frac{c{{x}^{3}}}{3} \right]_{0}^{1}\] |
| \[=a+\frac{b}{2}+\frac{c}{3}...(i)\] |
| Here, \[f(0)=a,f\left( \frac{1}{2} \right)=a+\frac{b}{2}+\frac{c}{4}\] |
| and \[f(1)=a+b+c\] |
| Now, \[\frac{f(0)+4f\left( \frac{1}{2} \right)+f(1)}{6}\] |
| \[=\frac{a+4\left( a+\frac{b}{2}+\frac{c}{4} \right)+a+b+c}{6}\] |
| \[=\frac{a+4\left( \frac{4a+2b+c}{4} \right)+a+b+c}{6}\] |
| \[=\frac{a+4a+2b+c+a+b+c}{6}=\frac{6a+3b+2c}{6}\] |
| \[=a+\frac{b}{2}+\frac{c}{3}\] |
| \[\therefore \] From equations (i) and (ii), we get |
| \[\int_{0}^{1}{f(x)dx=\frac{f(0)+4f\left( \frac{1}{2} \right)+f(1)}{6}}\] |
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