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

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

Since $f$ is a differentiable function, so extremum points of $f\left(x\right)$ we must have $f^{\mathrm{\prime }}\left(x\right)=0$ so $x=0,1/2$,1. Now $f^{\prime \prime }\left(x\right)=12x^2$

$-12x+2,f^{\prime \prime }\left(0\right)=2,f^{\prime \prime }\left(1\right)=2$ and $f^{\prime \prime }(1/2)=3-6+2=-1$

Thus the function has minimum at $x=0$ and $x=1$. Therefore, the required area $={\int }_0^1 \left(x^4-2x^3+x^2+3\right)\mathrm{d}x$ (See Chapter 14 for area as definite integral)

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