If f(x) is a quadratic in x,then int_{0}^{1} | vedclass

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Question

(A) Let $f(x) = ax^2 + bx + c$. Then,$\int_{0}^{1} f(x) dx = \int_{0}^{1} (ax^2 + bx + c) dx = \left[ \frac{ax^3}{3} + \frac{bx^2}{2} + cx \right]_0^1

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $f(x) = ax^2 + bx + c$. Then,$\int_{0}^{1} f(x) dx = \int_{0}^{1} (ax^2 + bx + c) dx = \left[ \frac{ax^3}{3} + \frac{bx^2}{2} + cx \right]_0^1 = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$. Now,calculate the values of $f(x)$ at the given points: $f(0) = c$ $f\left(\frac{1}{2}\right) = a\left(\frac{1}{4}\right) + b\left(\frac{1}{2}\right) + c = \frac{a + 2b + 4c}{4}$ $f(1) = a + b + c$ Consider the expression $\frac{1}{6}\left( f(0) + 4f\left(\frac{1}{2}\right) + f(1) \right)$: $= \frac{1}{6}\left( c + 4\left(\frac{a + 2b + 4c}{4}\right) + (a + b + c) \right)$ $= \frac{1}{6}\left( c + a + 2b + 4c + a + b + c \right)$ $= \frac{1}{6}\left( 2a + 3b + 6c \right)$. This matches the integral value. Thus,the correct option is $A$.

If $f(x)$ is a quadratic in $x$,then $\int_{0}^{1} f(x) dx$ is

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