Question

Let $f\left(x\right)=a+bx+c{x}^2,x>0,D<0$ then area bounded by coordinates axis line $x=2$ & $x=0$ is equals to

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

Answer 1

Answer: (D)

$\frac{1}{3}\left[f\right(0)+4f(1)+f(2\left)\right]$

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

Thus required area is, ${\int }_0^{2}f\left(x\right)dx$

$={\int }_0^2(a+bx+c{x}^2)dx=2a+2b+\frac{8}{3}c=\frac{1}{3}(6a+6b+8c)$ $\dots \left(1\right)$

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

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

$f\left(2\right)=a+2b+4c$

$f\left(0\right)+4f\left(1\right)+f\left(2\right)=a+4(a+b+c)+a+2b+4c=6a+6b+8c$

substitute in $\left(1\right)$

$\therefore {\int }_0^{2}f\left(x\right)dx=\frac{1}{3}\left[f\right(0)+4f(1)+f(2\left)\right]$