let the region r be the area enclosed by the function f(x)=ln(x) and g(x)=4/3x - 2. if the region r is the…

Explanation:

Step1: Find intersection points

Set $\ln(x)=\frac{4}{3}x - 2$. Using a calculator, the intersection points are $x = 1$ 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 $s=\ln(x)-(\frac{4}{3}x - 2)=\ln(x)-\frac{4}{3}x + 2$.

Step3: Set up volume integral

The volume $V$ of the solid with square cross - sections is given by the integral $V=\int_{a}^{b}s^{2}dx$, where $a = 1$, $b = 3$, and $s=\ln(x)-\frac{4}{3}x + 2$. So $V=\int_{1}^{3}(\ln(x)-\frac{4}{3}x + 2)^{2}dx$.

Step4: Expand the integrand

$(\ln(x)-\frac{4}{3}x + 2)^{2}=\ln^{2}(x)-\frac{8}{3}x\ln(x)+\frac{16}{9}x^{2}+4\ln(x)-\frac{16}{3}x + 4$.

Step5: Integrate term - by - term

$\int_{1}^{3}\ln^{2}(x)dx$, $\int_{1}^{3}-\frac{8}{3}x\ln(x)dx$, $\int_{1}^{3}\frac{16}{9}x^{2}dx$, $\int_{1}^{3}4\ln(x)dx$, $\int_{1}^{3}-\frac{16}{3}x dx$, $\int_{1}^{3}4dx$. Using integration by parts and integral formulas:

• $\int\ln^{2}(x)dx=x\ln^{2}(x)-2x\ln(x)+2x + C$
• $\int x\ln(x)dx=\frac{1}{2}x^{2}\ln(x)-\frac{1}{4}x^{2}+C$
• $\int x^{2}dx=\frac{1}{3}x^{3}+C$
• $\int\ln(x)dx=x\ln(x)-x + C$
• $\int xdx=\frac{1}{2}x^{2}+C$
• $\int dx=x + C$ Evaluating the definite integrals: $\int_{1}^{3}\ln^{2}(x)dx=[x\ln^{2}(x)-2x\ln(x)+2x]1^3=3\ln^{2}(3)-6\ln(3)+6-(0 - 0+2)=3\ln^{2}(3)-6\ln(3)+4$ $\int{1}^{3}-\frac{8}{3}x\ln(x)dx=-\frac{8}{3}[\frac{1}{2}x^{2}\ln(x)-\frac{1}{4}x^{2}]1^3=-\frac{8}{3}(\frac{9}{2}\ln(3)-\frac{9}{4}-\frac{1}{4})=-\frac{8}{3}(\frac{9}{2}\ln(3)-\frac{5}{2})=-12\ln(3)+\frac{20}{3}$ $\int{1}^{3}\frac{16}{9}x^{2}dx=\frac{16}{9}[\frac{1}{3}x^{3}]1^3=\frac{16}{9}(\frac{27}{3}-\frac{1}{3})=\frac{16}{9}\times\frac{26}{3}=\frac{416}{27}$ $\int{1}^{3}4\ln(x)dx=4[x\ln(x)-x]1^3=4(3\ln(3)-3-(0 - 1))=12\ln(3)-8$ $\int{1}^{3}-\frac{16}{3}x dx=-\frac{16}{3}[\frac{1}{2}x^{2}]1^3=-\frac{16}{3}(\frac{9}{2}-\frac{1}{2})=-\frac{16}{3}\times4=-\frac{64}{3}$ $\int{1}^{3}4dx=4[x]_1^3=4(3 - 1)=8$

Step6: Sum up the results

$V=(3\ln^{2}(3)-6\ln(3)+4)+(-12\ln(3)+\frac{20}{3})+\frac{416}{27}+(12\ln(3)-8)+(-\frac{64}{3})+8$ $V = 3\ln^{2}(3)-6\ln(3)-12\ln(3)+12\ln(3)+4+\frac{20}{3}+\frac{416}{27}-8-\frac{64}{3}+8$ $V = 3\ln^{2}(3)-6\ln(3)+\frac{108 + 180+416 - 576+216}{27}$ $V = 3\ln^{2}(3)-6\ln(3)+\frac{344}{27}\approx3\times(1.0986)^{2}-6\times1.0986+\frac{344}{27}$ $V\approx3\times1.2069 - 6.5916+\frac{344}{27}$ $V\approx3.6207-6.5916 + 12.7407$ $V\approx9.770$

Answer:

$9.770$