Question

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

• A. $\frac{91}{30}$ square units
• B. $\frac{93}{30}$ square units
• C. $\frac{1}{30}$ square units
• D. None of these

Answer 1

The correct option is A $\frac{91}{30}$ square units

The given curve is $y=x^4-2x^3+x^2+3,$ since $f$ is differentiable function

$\Rightarrow \frac{dy}{dx}=4x^3-6x^2+2x=2x(2x^2-3x+1)$

$\Rightarrow$ Thus $x=0,\frac{1}{2},1$ are critical points

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

$\Rightarrow$ Now taking the sign of $\frac{d^{2}y}{{dx}^2}$ at the critical point we get

${\left(\frac{d^{2}y}{{dx}^2}\right)}_{x=0}=2,{\left(\frac{d^{2}y}{{dx}^2}\right)}_{x=\frac{1}{2}}=-1,{\left(\frac{d^{2}y}{{dx}^2}\right)}_{x=1}=2$

Curve has minima at $x=0$ and $x=1:f(-\infty )=\infty ,f\left(0\right)=3,f\left(1\right)=3,f\left(\infty \right)=\infty$

$\therefore f\left(x\right)>0\forall xeR$

$\therefore$ Required area $={\int }_0^{1}ydx={\int }_0^1(x^4-2x^3+x^2+3)dx={[\frac{x^5}{5}-\frac{x^2}{2}+\frac{x^3}{3}+3x]}_0^1=\frac{91}{30}$ square units