Question

If $f\left(x\right)$ is quadratic in $x$, then ${\int }_0^{1}f\left(x\right)dx$ is

• A. $\frac{1}{6}\left(f\right(0)+4f(\frac{1}{2})+f(1\left)\right)$
• B. $\frac{1}{6}\left(4f\right(0)+f\left(\frac{1}{2}\right)+f(1\left)\right)$
• C. $\frac{1}{6}\left(f\right(0)+f(\frac{1}{2})+4f(1\left)\right)$
• D. $\frac{1}{6}\left(f\right(0)+f(\frac{1}{2})+f(1\left)\right)$

Answer 1

The correct option is A $\frac{1}{6}\left(f\right(0)+4f(\frac{1}{2})+f(1\left)\right)$

$f\left(x\right)=a{x}^2+bx+c$
${\int }_0^{1}f\left(x\right)dx=\frac{1}{6}(2a+3b+6c)$
$f\left(0\right)=c,f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c,f\left(1\right)=a+b+c$
$\Rightarrow f\left(0\right)+4f\left(\frac{1}{2}\right)+f\left(1\right)=2a+3b+6c$