Question

Find the length of the curve \(^{y = \frac{1}{6}{{(x2 + 4)}^{\frac{3}{2}}},0 \le x \le 3}\)

Short answer

The length of the curve is \(\frac{{15}}{2}\)

Step-by-step solution

Applied the length of the curve formula

By using the formula for length of the curve

\(\begin{aligned}L &= \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx\\b = 3\\a &= 0\\y &= \frac{1}{6}{({x^2} + 4)^{\frac{3}{2}}}\\\frac{{dy}}{{dx}} &= \left( {\frac{{x{{({x^2} + 4)}^{\frac{1}{2}}}}}{2}} \right)\end{aligned}\)

Solution of the length of the curve

\(\begin{aligned}L &= \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx\\ &= \int_0^3 {\sqrt {1 + {{\left( {\frac{{x{{({x^2} + 4)}^{\frac{1}{2}}}}}{2}} \right)}^2}} } dx\\ &= \frac{{15}}{2}\end{aligned}\)