let r be the region bounded by the functions f(x)=x² + 1 and g(x)=3x as shown in the diagram below. find the…

Explanation:

Step1: Find intersection points

Set $x^{2}+1 = 3x$, then $x^{2}-3x + 1=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=-3$, $c = 1$, we get $x=\frac{3\pm\sqrt{9 - 4}}{2}=\frac{3\pm\sqrt{5}}{2}$. Let $x_1=\frac{3-\sqrt{5}}{2}$ and $x_2=\frac{3+\sqrt{5}}{2}$.

Step2: Set up integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ is $A=\int_{x_1}^{x_2}|f(x)-g(x)|dx$. Here $f(x)=3x$ and $g(x)=x^{2}+1$, so $A=\int_{\frac{3 - \sqrt{5}}{2}}^{\frac{3+\sqrt{5}}{2}}(3x-(x^{2}+1))dx=\int_{\frac{3 - \sqrt{5}}{2}}^{\frac{3+\sqrt{5}}{2}}(-x^{2}+3x - 1)dx$.

Step3: Integrate

$\int(-x^{2}+3x - 1)dx=-\frac{1}{3}x^{3}+\frac{3}{2}x^{2}-x+C$.

Step4: Evaluate definite - integral

$A=\left[-\frac{1}{3}x^{3}+\frac{3}{2}x^{2}-x\right]_{\frac{3 - \sqrt{5}}{2}}^{\frac{3+\sqrt{5}}{2}}$. After substituting the upper and lower limits and simplifying (using a calculator), we get $A\approx2.323$.

Answer:

$2.323$