consider the following.\n$y = \\frac{2}{3}x^{3/2}+2$\nfind the arc length of the graph of the function over…

Explanation:

Step1: Find the derivative of $y$

Given $y = \frac{2}{3}x^{3/2}+2$, using the power - rule $(x^n)'=nx^{n - 1}$, we have $y'=\frac{2}{3}\times\frac{3}{2}x^{1/2}=x^{1/2}$.

Step2: Use the arc - length formula

The arc - length formula for a function $y = f(x)$ over the interval $[a,b]$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. Here, $(y')^{2}=x$, so the integral becomes $L=\int_{0}^{6}\sqrt{1 + x}dx$.

Step3: Use substitution

Let $u = 1 + x$, then $du=dx$. When $x = 0$, $u = 1$; when $x = 6$, $u = 7$. The integral $\int_{0}^{6}\sqrt{1 + x}dx=\int_{1}^{7}u^{1/2}du$.

Step4: Integrate $u^{1/2}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int_{1}^{7}u^{1/2}du=\left[\frac{2}{3}u^{3/2}\right]_{1}^{7}$.

Step5: Evaluate the definite integral

$\frac{2}{3}u^{3/2}\big|_{1}^{7}=\frac{2}{3}(7^{3/2}-1^{3/2})=\frac{2}{3}(7\sqrt{7}-1)\approx\frac{2}{3}(7\times2.646 - 1)=\frac{2}{3}(18.522-1)=\frac{2}{3}\times17.522\approx11.681$.

Answer:

$11.681$