Question

If f(x) $=4x^3+3x^2+2x+1$ then area bounded by $x=0,y=0$ and $x=2$ is

• A. $30$
• B. $20$
• C. $25$
• D. $34$

Answer 1

The correct option is A $30$

Here f' (x)$=12x^2+6x+2$
For x $ϵ$ $(0,2)$
$f^{\prime }\left(x\right)>0$, then $f\left(x\right)\ge 0$
So $A={\int }_0^2(4x^3+3x^2+2x+1)dx$
$A={[x^4+x^3+x^2+x]}_0^2$
$=16+8+4+2$
$=30$ sq.units