Question

If $f\left(x\right)=a+bx+c{x}^2$, show that
$\underset{0}{\stackrel{1}{\int }}f\left(x\right)dx=\frac{1}{6}\left[f\right(0)+4f\left(\frac{1}{2}\right)+f(1\left)\right]$

Answer 1

$f\left(x\right)=a+bx+c{x}^2$

$\therefore f\left(0\right)=a$

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

$\therefore f\left(1\right)=a+b+c$

$\therefore {\int }_0^{1}f\left(x\right)dx={\int }_0^1(a+bx+c{x}^2)dx$

$=ax+\frac{b{x}^2}{2}+\frac{c{x}^3}{3}{|}_0^1$

$=a(1-0)+\frac{b}{2}(1-0)+\frac{c}{3}(1-0)$

$=a+\frac{b}{2}+\frac{c}{3}......\left(1\right)\to LHS$

$\therefore RHS=\frac{1}{6}\left[f\right(0)+4f(1/2)+f(1\left)\right]$

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

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

$=a+\frac{3b}{6}+\frac{2c}{6}$

$=a+\frac{b}{2}+\frac{c}{3}\to RHS$

$\therefore LHS=RHS$

Hence proved.