The area bounded by y = f(x) = x⁴ – 2x³ + x² + 3, X-axis and ordinates corresponding to minimum of the function f(x) is __ sq.units

$\mathrm{y} = \text{⨍ }\left(\mathrm{x}\right) = {\mathrm{x}}^4–2{\mathrm{x}}^3+\mathrm{x}2+3,$

$f^{\mathrm{\prime }}\left(x\right)=4x^3-6x^2+2x=0\Rightarrow 2x(x-1)(2x-1)=0\Rightarrow x=0,1,1/2$

$f^{\prime \prime }\left(x\right)=12x^2-12x+2=2\left(6x^2-6x+1\right)$

${\mathrm{f}}^{\prime \prime }(\mathrm{x}{)}_{x=0}>0\Rightarrow \text{ Minimum exists at }\mathrm{x}=0$

${\mathrm{f}}^{\prime \prime }(\mathrm{x}{)}_{x=1/2}>0\Rightarrow \text{ Minimum exists at }\mathrm{x}=1/2$

${\mathrm{f}}^{\prime \prime }(\mathrm{x}{)}_{x=1}>0\Rightarrow \text{ Minimum exists at }\mathrm{x}=1$

$\mathbf{\Rightarrow }\mathbf{\text{ The curve is bounded by the ordinates }}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{\text{ and }}\mathit{x}\mathbf{=}\mathbf{1}$

$\therefore \text{ Required Area }{\int }_0^1 \left(x^4-2x^3+x^2+3\right)\mathrm{dx}$

${\left.\left(\frac{x^5}{5}-\frac{x^4}{2}+\frac{x^3}{3}+3x\right)\right|}_0^1=\frac{91}{30}\text{ sq. units }$