MCQ
If $f(x)$ is a quadratic in $x$ , then $\int\limits_0^1 {f(x) dx}$ is
  • $\frac{1}{6}\left( {f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f(1)} \right)$
  • B
    $\frac{1}{6}\left( {4f\left( 0 \right) + f\left( {\frac{1}{2}} \right) + f(1)} \right)$
  • C
    $\frac{1}{6}\left( {f\left( 0 \right) + f\left( {\frac{1}{2}} \right) + 4f(1)} \right)$
  • D
    $\frac{1}{6}\left( {f\left( 0 \right) + f\left( {\frac{1}{2}} \right) + f(1)} \right)$

Answer

Correct option: A.
$\frac{1}{6}\left( {f\left( 0 \right) + 4f\left( {\frac{1}{2}} \right) + f(1)} \right)$
a
If $f(x)=a x^{2}+b x+c$

$\int_{0}^{1} f(x) d x=\frac{1}{6}(2 a+3 b+6 c)$

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

$f(1)=a+b+c$

$\Rightarrow f(0)+4 f\left(\frac{1}{2}\right)+f(1)=2 \mathrm{a}+3 \mathrm{b}+6 \mathrm{c}$

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