Mathematics - Area Under Curves Question with Solution | TestHub

The curve is $y=|x^2-4x+3|$. The roots of $x^2-4x+3=0$ are $x=1$ and $x=3$. The vertex of $y=x^2-4x+3$ is at $(2,-1)$. The curve $y=|x^2-4x+3|$ is symmetric about $x=2$. We find the intersection points of $y=|x^2-4x+3|$ and $y=3$. This occurs when $x^2-4x+3=3$ or $x^2-4x+3=-3$. Case 1: $x^2-4x+3=3⟹x^2-4x=0⟹x(x-4)=0⟹x=0,x=4$. Case 2: $x^2-4x+3=-3⟹x^2-4x+6=0$. The discriminant is $(-4{)}^2-4(1)(6)=-8<0$, so there are no real solutions.

The region $R_1$ is bounded by $y=3$ from above and $y=|x^2-4x+3|$ from below, between $x=0$ and $x=4$. The area is given by $A={\int }_0^4(3-|x^2-4x+3|)dx$. We split the integral based on the definition of $|x^2-4x+3|$:$A={\int }_0^1(3-(x^2-4x+3))dx+{\int }_1^3(3-(-(x^2-4x+3)))dx+{\int }_3^4(3-(x^2-4x+3))dx$ $A={\int }_0^1(-x^2+4x)dx+{\int }_1^3(x^2-4x+6)dx+{\int }_3^4(-x^2+4x)dx$ Calculating each part: 1. $[-\frac{x^3}{3}+2x^2{]}_0^1=(-\frac{1}{3}+2)-0=\frac{5}{3}$. 2. $[\frac{x^3}{3}-2x^2+6x{]}_1^3=(9-18+18)-(\frac{1}{3}-2+6)=9-(\frac{13}{3})=\frac{14}{3}$. 3. $[-\frac{x^3}{3}+2x^2{]}_3^4=(-\frac{64}{3}+32)-(-\frac{27}{3}+18)=(\frac{32}{3})-9=\frac{5}{3}$. Total Area $=\frac{5}{3}+\frac{14}{3}+\frac{5}{3}=\frac{24}{3}=8$.