Question

The area bounded by the curve $y=x^4-2x^3+x^2+3,$ the axis of abscissas and two ordinates corresponding to the points of minimum of the function $y\left(x\right)$ is:

• A. $\frac{10}{3}$ sq.unit
• B. $\frac{27}{10}$ sq.unit
• C. $\frac{21}{10}$ sq.unit
• D. $\frac{91}{30}$ sq.unit.

Answer 1

The correct option is D $\frac{91}{30}$ sq.unit.

Given curve is
$y=x^4-2x^3+x^2+3$
$\therefore \frac{dy}{dx}=4x^3-6x^2+2x$
$\frac{d^{2}y}{d{x}^2}=12x^2-12x+2$
For maxima and minima $\frac{dy}{dx}=0$
$\therefore 4x^3-6x^2+2x=0$
$\Rightarrow 2x(2x^2-3x+1)=0$
$\therefore x=0,\frac{1}{2},1$
$\therefore {\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=0}=2(6x^2-6x+1)$
$\therefore {\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=0}=2,{\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=\frac{1}{2}}=-1$ and ${\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=1}=2$
$\therefore$ Points of minimum are $x=0$ and $x=1$
$\therefore$ Required Area$={\int }_0^1(x^4-2x^3+x^2+3)dx$
$={[\frac{x^5}{5}-\frac{{2x}^4}{4}+\frac{x^2}{2}+3x]}_0^1$
$=\frac{1}{5}-\frac{2}{4}+\frac{1}{3}+3$
$=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3=\frac{91}{30}$