question let the region r be the area enclosed by the function f(x)=x² - 2 and g(x)=3x - 2. if the region r…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$, so $x^{2}-2 = 3x - 2$. Rearranging gives $x^{2}-3x=0$, factoring out $x$ we get $x(x - 3)=0$. The solutions are $x = 0$ and $x=3$.

Step2: Determine side - length of square cross - section

The side - length $s$ of each square cross - section perpendicular to the $x$ - axis is $s=(3x - 2)-(x^{2}-2)=3x - x^{2}$.

Step3: Find the volume formula

The volume $V$ of the solid with square cross - sections is given by the integral $V=\int_{a}^{b}s^{2}dx$, where $a = 0$, $b = 3$ and $s = 3x - x^{2}$. So $V=\int_{0}^{3}(3x - x^{2})^{2}dx$.

Step4: Expand the integrand

Expand $(3x - x^{2})^{2}=(3x - x^{2})(3x - x^{2})=9x^{2}-6x^{3}+x^{4}$.

Step5: Integrate term - by - term

$\int(9x^{2}-6x^{3}+x^{4})dx=9\times\frac{x^{3}}{3}-6\times\frac{x^{4}}{4}+\frac{x^{5}}{5}+C = 3x^{3}-\frac{3}{2}x^{4}+\frac{1}{5}x^{5}+C$.

Step6: Evaluate the definite integral

$V=\left[3x^{3}-\frac{3}{2}x^{4}+\frac{1}{5}x^{5}\right]_{0}^{3}=3\times3^{3}-\frac{3}{2}\times3^{4}+\frac{1}{5}\times3^{5}$. $V = 3\times27-\frac{3}{2}\times81+\frac{1}{5}\times243$. $V = 81-\frac{243}{2}+\frac{243}{5}$. $V=\frac{810 - 1215+486}{10}=\frac{71}{10}=7.125$.

Answer:

$7.125$