$If for all real triplets (a, b, c), f (x) = a + b x + c x^{2}; then \int_{0}^{1} f (x) d x is equal to:$

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$\frac{1}{2} \{f (1) + 3 f (\frac{1}{2})\}$

$\frac{1}{3} \{f (0) + f (\frac{1}{2})\}$

$\frac{1}{6} \{f (0) + f (1) + 4 f (\frac{1}{2})\}$

$2 \{3 f (1) + 2 f (\frac{1}{2})\}$

answer is D.

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Detailed Solution

$\int_{0}^{1} (a + b x + c x^{2}) d x = {⌊a x + \frac{b x^{2}}{2} + \frac{c x^{3}}{3}]}_{0}^{1} = a + \frac{b}{2} + \frac{c}{3}$

$\begin{array}{l} We have f (1) = a + b + c, f (0) = a, f (\frac{1}{2}) = a + \frac{b}{2} + \frac{c}{4} \\ Now \end{array}$

$\frac{1}{6} (f (1) + f (0) + 4 f (\frac{1}{2})) = \frac{1}{6} (a + b + c + a + 4 (a + \frac{b}{2} + \frac{c}{4}))$

$\begin{array}{l} = \frac{1}{6} (6 a + 3 b + 2 c) \\ = a + \frac{b}{2} + \frac{c}{3} \end{array}$

Therefore, it is matching with the option 4

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