Question

The area bounded by the curve $y=x^4-2x^3+x^2+3$ with x-axis and ordinates corresponding to the minima of y is :

• A. 1
• B. $\frac{91}{30}$
• C. $\frac{30}{9}$
• D. 4

Answer 1

The correct option is B $\frac{91}{30}$

The given curve is $y=x^4-2x^3+x^2+3$ ……..$\left(1\right)$

To find max on min points we got,

$\frac{dy}{dx}=4x^3-6x^2+2x$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=12x^2-12x+2$

For max or min

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

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

$2x(2x^2-3x+1)=0$

$x=0$ or $2x^2-3x+1=0$

$2x^2-3x+1=0$

$(2x-1)(x-1)=0$

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

Now, $\left[\frac{d^{2}y}{d{x}^2}\right]x=0=2>0$

So, $f\left(x\right)$ has a min value at $x=2$

${\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=\frac{1}{2}}=12.\frac{1}{4}-12.\frac{1}{2}+2$

$=-3-6+2=-1<0$

So $f\left(x\right)$ has min value at $x=\frac{1}{2}$

Finally ${\left[\frac{d^{2}y}{d{x}^2}\right]}_{x=1}=12-12+2=2>0$

So, $f\left(x\right)$ has minimum value at $x=1$

Hence the required area bounded by the curve $\left(1\right)$, the x-axis and the $2$ line $x=0$ & $x=1$ is

${\int }_0^{1}ydx$

$={\int }_0^1[x^2-2x^3+x^2+3]dx$

$=[\frac{x^5}{5}-2.\frac{x^4}{4}+\frac{x^3}{3}+3x]x=1$ $x=0$

$=\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3-0$

$=\frac{6-15+10+90}{30}$

$=\frac{91}{30}$.