Question

If for all real triplets (a, b, c), $f(x ) = a + bx + cx^2$; then $\int\limits^{1}_{0}$ 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)

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