Mathematics - Area Under Curves Question with Solution | TestHub

The area function is $A(k)={\int }_k^{k+1}|x^2-4x+3|dx$. We need to find the minimum of $A(k)$ for $k\in [0,3]$. By Leibniz integral rule, $A^{\prime }(k)=|f(k+1)|-|f(k)|$. For critical points, $A^{\prime }(k)=0⟹|f(k+1)|=|f(k)|$. This implies either $f(k+1)=f(k)$ or $f(k+1)=-f(k)$. Case 1: $f(k+1)=f(k)⟹(k+1{)}^2-4(k+1)+3=k^2-4k+3⟹k^2-2k=k^2-4k+3⟹2k=3⟹k=3/2$. This is a critical point. $A(3/2)={\int }_{3/2}^{5/2}-(x^2-4x+3)dx=[-\frac{x^3}{3}+2x^2-3x{]}_{3/2}^{5/2}=\frac{11}{12}$. $A^{\prime \prime }(3/2)=-2<0$, so this is a local maximum. Case 2: $f(k+1)=-f(k)⟹(k+1{)}^2-4(k+1)+3=-(k^2-4k+3)⟹2k^2-6k+3=0$. The solutions are $k_1=\frac{3-\sqrt{3}}{2}$ and $k_2=\frac{3+\sqrt{3}}{2}$. Both are in $[0,3]$. Let $F(x)=\int (x^2-4x+3)dx=\frac{x^3}{3}-2x^2+3x$. For $x$ such that $2x^2-6x+3=0$, $F(x)=\frac{3-x}{2}$. For $k_1=\frac{3-\sqrt{3}}{2}\approx 0.634$: The interval $[k_1,k_1+1]$ crosses $x=1$. $A(k_1)={\int }_{k_1}^{1}f(x)dx+{\int }_1^{k_1+1}-f(x)dx=(F(1)-F(k_1))-(F(k_1+1)-F(1))=2F(1)-F(k_1)-F(k_1+1)$.$F(1)=4/3$. $F(k_1)=\frac{3-k_1}{2}=\frac{3-(3-\sqrt{3})/2}{2}=\frac{3+\sqrt{3}}{4}$. $F(k_1+1)=\frac{3-(k_1+1)}{2}=\frac{2-k_1}{2}=\frac{1+\sqrt{3}}{4}$.$A(k_1)=2(\frac{4}{3})-\frac{3+\sqrt{3}}{4}-\frac{1+\sqrt{3}}{4}=\frac{8}{3}-\frac{4+2\sqrt{3}}{4}=\frac{8}{3}-\frac{2+\sqrt{3}}{2}=\frac{16-3(2+\sqrt{3})}{6}=\frac{10-3\sqrt{3}}{6}\approx 0.800$. For $k_2=\frac{3+\sqrt{3}}{2}\approx 2.366$: The interval $[k_2,k_2+1]$ crosses $x=3$. $A(k_2)={\int }_{k_2}^3-f(x)dx+{\int }_3^{k_2+1}f(x)dx=(-F(3)+F(k_2))+(F(k_2+1)-F(3))$. Since $F(3)=0$, $A(k_2)=F(k_2)+F(k_2+1)$.$F(k_2)=\frac{3-k_2}{2}=\frac{3-(3+\sqrt{3})/2}{2}=\frac{3-\sqrt{3}}{4}$. $F(k_2+1)=\frac{3-(k_2+1)}{2}=\frac{2-k_2}{2}=\frac{1-\sqrt{3}}{4}$.$A(k_2)=\frac{3-\sqrt{3}}{4}+\frac{1-\sqrt{3}}{4}=\frac{4-2\sqrt{3}}{4}=\frac{2-\sqrt{3}}{2}\approx 0.134$. Comparing values at critical points and endpoints:$A(0)={\int }_0^1(x^2-4x+3)dx=[\frac{x^3}{3}-2x^2+3x{]}_0^1=\frac{4}{3}\approx 1.333$.$A(3)={\int }_3^4(x^2-4x+3)dx=[\frac{x^3}{3}-2x^2+3x{]}_3^4=\frac{4}{3}\approx 1.333$.$A(1)={\int }_1^2-(x^2-4x+3)dx=[-\frac{x^3}{3}+2x^2-3x{]}_1^2=\frac{2}{3}\approx 0.667$.$A(2)={\int }_2^3-(x^2-4x+3)dx=[-\frac{x^3}{3}+2x^2-3x{]}_2^3=\frac{2}{3}\approx 0.667$. The minimum value among these is $A(k_2)=\frac{2-\sqrt{3}}{2}\approx 0.134$.