Question

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

• A. $2\left\{3f\left(1\right)+2f\left(\frac{1}{2}\right)\right\}$
• B. $\frac{1}{3}\left\{f\left(0\right)+f\left(\frac{1}{2}\right)\right\}$
• C. $\frac{1}{2}\left\{f\left(1\right)+3f\left(\frac{1}{2}\right)\right\}$
• D. $\frac{1}{6}\left\{f\left(0\right)+f\left(1\right)+4f\left(\frac{1}{2}\right)\right\}$

Answer 1

Answer: (D)

$ƒ\left(x\right)=a+bx+c{x}^2$
$\underset{0}{\overset{1}{\int }}f(x)dx$${\left[ax+\frac{b{x}^2}{2}+\frac{c{x}^3}{3}\right]}_{{}_{{}_0}}^{{}^1}$
$=a+\frac{b}{2}+\frac{c}{3}=\frac{1}{6}\left[6a+3b+c\right]$
$=\frac{1}{6}\left[f\left(0\right)+f\left(1\right)+4f\left(\frac{1}{2}\right)\right]$