3. find the arclength of $y = 6xsqrt{x}$ on $1leq xleq3$.

Explanation:

Step1: Rewrite the function

First, rewrite $y = 6x\sqrt{x}=6x^{\frac{3}{2}}$. Then find its derivative $y'$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $y'=6\times\frac{3}{2}x^{\frac{3}{2}-1}=9x^{\frac{1}{2}} = 9\sqrt{x}$.

Step2: Use the arc - length formula

The arc - length formula for a function $y = f(x)$ from $x=a$ to $x = b$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. Here, $a = 1$, $b = 3$, and $(y')^{2}=(9\sqrt{x})^{2}=81x$. So, $L=\int_{1}^{3}\sqrt{1 + 81x}dx$.

Step3: Use substitution

Let $u=1 + 81x$, then $du=81dx$. When $x = 1$, $u=1+81\times1 = 82$; when $x = 3$, $u=1+81\times3=244$. The integral becomes $\frac{1}{81}\int_{82}^{244}\sqrt{u}du=\frac{1}{81}\int_{82}^{244}u^{\frac{1}{2}}du$.

Step4: Integrate

Integrating $u^{\frac{1}{2}}$ using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we get $\frac{1}{81}\times\frac{2}{3}u^{\frac{3}{2}}\big|{82}^{244}=\frac{2}{243}(u^{\frac{3}{2}})\big|{82}^{244}$.

Step5: Evaluate the definite integral

$\frac{2}{243}(244^{\frac{3}{2}}-82^{\frac{3}{2}})=\frac{2}{243}(244\sqrt{244}-82\sqrt{82})=\frac{2}{243}(244\times2\sqrt{61}-82\sqrt{82})=\frac{4}{243}(122\sqrt{61}-41\sqrt{82})$.

Answer:

$\frac{2}{243}(244^{\frac{3}{2}}-82^{\frac{3}{2}})$