find the area of the region bounded by the graphs of the given equations. y = x^2 - 2x + 1, y = x + 1

Explanation:

Step1: Find intersection points

Set $x^{2}-2x + 1=x + 1$. $x^{2}-3x=0$. Factor out $x$: $x(x - 3)=0$. So $x = 0$ and $x=3$.

Step2: Determine upper - lower functions

For $0\leq x\leq3$, $y=x + 1$ is above $y=x^{2}-2x + 1$. The area $A=\int_{a}^{b}[(upper\ function)-(lower\ function)]dx$. Here $a = 0$, $b = 3$, upper function is $x + 1$, lower function is $x^{2}-2x + 1$. So $A=\int_{0}^{3}[(x + 1)-(x^{2}-2x + 1)]dx$. Simplify the integrand: $(x + 1)-(x^{2}-2x + 1)=x + 1-x^{2}+2x - 1=-x^{2}+3x$.

Step3: Evaluate the integral

$A=\int_{0}^{3}(-x^{2}+3x)dx$. Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have: $\int(-x^{2}+3x)dx=-\frac{x^{3}}{3}+\frac{3x^{2}}{2}+C$. $A=\left[-\frac{x^{3}}{3}+\frac{3x^{2}}{2}\right]_{0}^{3}$. $A=\left(-\frac{3^{3}}{3}+\frac{3\times3^{2}}{2}\right)-\left(-\frac{0^{3}}{3}+\frac{3\times0^{2}}{2}\right)$. $A=\left(-9+\frac{27}{2}\right)-0=\frac{-18 + 27}{2}=\frac{9}{2}$.

Answer:

$\frac{9}{2}$