Question

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

• A. $\frac{1}{30}$
• B. $\frac{1}{10}$
• C. $\frac{91}{30}$
• D. $3$

Answer 1

Answer: (C)

$\frac{91}{30}$

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 and minima $\frac{dy}{dx}=0$

$\therefore 4x^3-6x^2+2=0$

$\Rightarrow x=0,1,\frac{1}{2}$

$\therefore \frac{d^{2}y}{d{x}_{x=0}^2}=2,\frac{d^{2}y}{d{x}_{x=\frac{1}{2}}^2}=-1$ and $\frac{d^{2}y}{d{x}_{x=1}^2}=2$

Therefore point of minima are $x=0$ and $x=1$

Therefore required area $={\int }_0^1(x^4-2x^3+x^2+3)dx$

$={(\frac{x^5}{5}-2\frac{x^4}{4}+\frac{x^3}{3}+3x)}_0^1=\frac{91}{30}$.