The equation of given curve is $ y={{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+3 $ On differentiating w. r. t. $ x, $ we get $ \frac{dy}{dx}=4{{x}^{3}}-6{{x}^{2}}+2x $ Again differentiating, we get $ \frac{{{d}^{2}}y}{d{{x}^{2}}}=12{{x}^{2}}-12x+2 $ Put $ \frac{dy}{dx}=0 $ for maxima or minima $ \Rightarrow $ $ 2x(2{{x}^{2}}-3x+1)=0 $ $ \Rightarrow $ $ x=0,1,\frac{1}{2} $ $ \therefore $ $ {{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=0}}=2 $ $ \therefore $ Function is minimum at $ x=0 $ and $ {{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=1}}=12-12+2=2 $ $ \therefore $ Function is minimum at $ x=\frac{1}{2} $ . also $ {{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=\frac{1}{2}}}=-1<0 $ $ \therefore $ Function is maximum at $ x=\frac{1}{2} $ . $ \therefore $ Required area $ =\int_{0}^{1}{({{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+3})dx $ $ =\left[ \frac{{{x}^{5}}}{5}-\frac{2{{x}^{4}}}{4}+\frac{{{x}^{3}}}{3}+3x \right]_{0}^{1} $ $ =\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3=\frac{91}{30}\,sq\,unit $