The arc length of a function is given by the equation:
[tex]L\text{ = }\int ^b_a\sqrt[]{1+\mleft(f\prime(x)\mright)^2}dx[/tex]f'(x) = 3x^(1/2)
Now, applying the arc length formula:
[tex]\begin{gathered} L\text{ = }\int ^3_0\sqrt[]{1+(3x^{\frac{1}{2}})^2}dx \\ L\text{ = }\int ^3_0\sqrt[]{1+9x}dx;\text{ }substituting\text{ t = 1 + 9x} \\ L\text{ = }\frac{1}{9}\int ^3_0\sqrt[]{t}\text{ dt} \\ L\text{ = }\frac{1}{9}\times\frac{2\sqrt[]{t}\text{ }\times\text{ |t|}}{3};\text{ }replacing\colon \\ L\text{ = }\frac{1}{9}\times\frac{2\sqrt[]{1+9x}\text{ }\times\text{ |1+9x|}}{3};\text{ from 0 to 3; evaluating the limits:} \\ L\text{ = }\frac{112\sqrt[]{7}-2}{27}=\text{ }10.90 \end{gathered}[/tex]
