Question

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

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

Answer 1

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

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

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