this question is designed to be answered with a calculator. the region bounded by y = x² - 3x and y = 3 - x…

Answer:

C. 107.233

Explanation:

Step1: Find intersection points

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

Step2: Use the washer - method formula

The outer radius $R=4-(x^{2}-3x)$ and the inner radius $r = 4-(3 - x)$. The volume $V=\pi\int_{a}^{b}(R^{2}-r^{2})dx=\pi\int_{-1}^{3}[(4-(x^{2}-3x))^{2}-(4-(3 - x))^{2}]dx$.

Step3: Expand and integrate

Expand the integrand: $(4-(x^{2}-3x))^{2}-(4-(3 - x))^{2}=(4 - x^{2}+3x)^{2}-(1 + x)^{2}$. $(4 - x^{2}+3x)^{2}=16-8x^{2}+6x + x^{4}-6x^{3}+9x^{2}=x^{4}-6x^{3}+x^{2}+6x + 16$. $(1 + x)^{2}=1 + 2x+x^{2}$. The integrand is $x^{4}-6x^{3}+6x + 15$. Integrate term - by - term: $\int(x^{4}-6x^{3}+6x + 15)dx=\frac{1}{5}x^{5}-\frac{3}{2}x^{4}+3x^{2}+15x+C$.

Step4: Evaluate the definite integral

$V=\pi\left[\frac{1}{5}x^{5}-\frac{3}{2}x^{4}+3x^{2}+15x\right]_{-1}^{3}$. $V=\pi\left[\left(\frac{1}{5}(3)^{5}-\frac{3}{2}(3)^{4}+3(3)^{2}+15(3)\right)-\left(\frac{1}{5}(-1)^{5}-\frac{3}{2}(-1)^{4}+3(-1)^{2}+15(-1)\right)\right]$. $V=\pi\left[\left(\frac{243}{5}-\frac{243}{2}+27 + 45\right)-\left(-\frac{1}{5}-\frac{3}{2}+3-15\right)\right]$. $V\approx107.233$.