4. the area of the region enclosed by the graphs of y=x and $y = x^{2}-3x + 3$ is (a) $\frac{2}{3}$ (b) 1…

4. the area of the region enclosed by the graphs of y=x and $y = x^{2}-3x + 3$ is (a) $\frac{2}{3}$ (b) 1 (c) $\frac{4}{3}$ (d) 2 (e) $\frac{14}{3}$

4. the area of the region enclosed by the graphs of y=x and $y = x^{2}-3x + 3$ is (a) $\frac{2}{3}$ (b) 1 (c) $\frac{4}{3}$ (d) 2 (e) $\frac{14}{3}$

Answer

Explanation:

Step1: Find intersection points

Set (x = x^{2}-3x + 3), then (x^{2}-4x + 3=0). Factor: ((x - 1)(x - 3)=0). So (x=1) or (x = 3).

Step2: Set up the integral

The area (A=\int_{a}^{b}\left|f(x)-g(x)\right|dx). Here (f(x)=x), (g(x)=x^{2}-3x + 3), (a = 1), (b = 3). Since (x\geq x^{2}-3x + 3) for (1\leq x\leq3), (A=\int_{1}^{3}\left[x-(x^{2}-3x + 3)\right]dx=\int_{1}^{3}(-x^{2}+4x - 3)dx).

Step3: Integrate

(\int(-x^{2}+4x - 3)dx=-\frac{1}{3}x^{3}+2x^{2}-3x+C). Evaluate from (1) to (3): (\left(-\frac{1}{3}(3)^{3}+2(3)^{2}-3(3)\right)-\left(-\frac{1}{3}(1)^{3}+2(1)^{2}-3(1)\right)) (=( - 9 + 18-9)-\left(-\frac{1}{3}+2 - 3\right)) (=0-\left(-\frac{1}{3}-1\right)= \frac{4}{3})

Answer:

C. (\frac{4}{3})