If for all real triplets (a, b, c),f(x) = a + | vedclass

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(C) Given $f(x) = a + bx + cx^2$. Calculating the definite integral: $\int_{0}^{1} (a + bx + cx^2) dx = [ax + \frac{bx^2}{2} + \frac{cx^3}{3}]_{0}^{1}

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

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(C) Given $f(x) = a + bx + cx^2$. Calculating the definite integral: $\int_{0}^{1} (a + bx + cx^2) dx = [ax + \frac{bx^2}{2} + \frac{cx^3}{3}]_{0}^{1} = a + \frac{b}{2} + \frac{c}{3}$. Simplifying the expression: $a + \frac{b}{2} + \frac{c}{3} = \frac{6a + 3b + 2c}{6}$. Now,evaluate the terms in option $C$: $f(0) = a$. $f(1) = a + b + c$. $f(\frac{1}{2}) = a + \frac{b}{2} + \frac{c}{4}$. Substituting these into $\frac{1}{6} \{f(0) + f(1) + 4f(\frac{1}{2})\}$: $= \frac{1}{6} \{a + (a + b + c) + 4(a + \frac{b}{2} + \frac{c}{4})\}$ $= \frac{1}{6} \{a + a + b + c + 4a + 2b + c\}$ $= \frac{1}{6} \{6a + 3b + 2c\} = a + \frac{b}{2} + \frac{c}{3}$. Thus,the correct option is $C$.

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

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