Question

Let $x=x_1,x=x_2$ be two points of minima of $y=x^4-2x^3+x^2+3.$ The area bounded by the curve, x - axis and $x=x_1,x=x_2$ in sq. units is:

• A. $\frac{81}{30}$
• B. $\frac{91}{30}$
• C. $\frac{71}{30}$
• D. None of the above

Answer 1

Answer: (B)

$\frac{91}{30}$

$y=x^4-2x^3+x^2+3$
$\frac{dy}{dx}=4x^3-6x^2+2x$
For maxima or minima,
$\frac{dy}{dx}=0$
$\Rightarrow 2x(2x^2-3x+1)=0$
$\Rightarrow 2x(x-1)(2x-1)=0$
$\Rightarrow x=0,1,\frac{1}{2}$
Now, $\frac{d^{2}y}{d{x}^2}=12x^2-12x+2$
At $x=0$, $\frac{d^{2}y}{d{x}^2}>0$
So, $x=0$ is point of minima.
At $x=\frac{1}{2}$ ,$\frac{d^{2}y}{d{x}^2}<0$
So, $x=\frac{1}{2}$ is point of maxima.
At $x=1$, $\frac{d^{2}y}{d{x}^2}>0$
So, $x=1$ is point of minima.
Hence, $x_1=0$ and $x_2=1$
Required area $={\int }_0^{1}ydx$
$={\int }_0^1(x^4-2x^3+x^2+3)dx$
$=[\frac{x^5}{5}-\frac{x^4}{2}+\frac{x^3}{3}+3x{]}_0^1$
$=[\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3]$
$=\frac{91}{30}$sq. units