find the area of each of the region bounded of the following curves. write your solutions on the space…

Explanation:

Step1: Find intersection points

Set $x^{2}-6x + 7=x - 1$. Rearrange to get $x^{2}-7x + 8=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=-7$, $c = 8$, we have $x=\frac{7\pm\sqrt{49 - 32}}{2}=\frac{7\pm\sqrt{17}}{2}$.

Step2: Determine the upper - lower functions

For $\frac{7-\sqrt{17}}{2}<x<\frac{7 + \sqrt{17}}{2}$, we check a value in the interval, say $x = 3$. $y_1=3^{2}-6\times3 + 7=9-18 + 7=-2$ and $y_2=3 - 1=2$. So $y=x - 1$ is the upper - function and $y=x^{2}-6x + 7$ is the lower - function.

Step3: Calculate the area

The area $A=\int_{\frac{7-\sqrt{17}}{2}}^{\frac{7+\sqrt{17}}{2}}[(x - 1)-(x^{2}-6x + 7)]dx$. Simplify the integrand: $(x - 1)-(x^{2}-6x + 7)=-x^{2}+7x - 8$. Integrate: $\int(-x^{2}+7x - 8)dx=-\frac{1}{3}x^{3}+\frac{7}{2}x^{2}-8x+C$. Evaluate the definite integral: [ \begin{align*} &\left(-\frac{1}{3}\left(\frac{7+\sqrt{17}}{2}\right)^{3}+\frac{7}{2}\left(\frac{7+\sqrt{17}}{2}\right)^{2}-8\left(\frac{7+\sqrt{17}}{2}\right)\right)-\left(-\frac{1}{3}\left(\frac{7-\sqrt{17}}{2}\right)^{3}+\frac{7}{2}\left(\frac{7-\sqrt{17}}{2}\right)^{2}-8\left(\frac{7-\sqrt{17}}{2}\right)\right)\ =&\frac{17\sqrt{17}}{6} \end{align*} ]

Answer:

$\frac{17\sqrt{17}}{6}$