Question

The area bounded by $y=x^4-2x^3+x^2+3$, axis of abscissa and two ordinates corresponding two points of minima of the function $y=f\left(x\right)$ is

• A. $1/30$
• B. $1/10$
• C. $91/30$
• D. $3$

Answer 1

Answer: (C)

$91/30$

The given curve is $y=x^4-2x^3+x^2+3$
$\therefore \frac{dy}{dx}=4x^3-6x^2+2x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=12x^2-12+2$
For maxima nad minima $\frac{dy}{dx}=0$
$\therefore 4x^3-6x^2+2x=0$
$\Rightarrow x=0,1,\frac{1}{2}$
$\therefore {\frac{d^{2}y}{d{x}^2}}_{x=0}=2,{\frac{d^{2}y}{d{x}^2}}_{x=\frac{1}{2}}=-1$ and ${\frac{d^{2}y}{d{x}^2}}_{x=1}=2$
Therefore point of minima are $x=0$ and $x=1$
Therefore required area $={\int }_0^1(x^4-2x^3+x^2+3)dx$
$={(\frac{x^5}{5}-2\frac{x^4}{4}+\frac{x^3}{3}+3x)}_0^1=\frac{91}{30}$