Find the length of the curve. Select the correct answer.
$$y = \frac { 1 } { 6 } \left( x ^ { 2 } + 4 \right) ^ { 3 / 2 } , 0 \leq x \leq 4$$
A) $\quad 14.0000$
B) $\quad 14.6667$
C) $15.3333$
D) $14.3333$
E) $13.6667$

Question

Quizplus · Accepted Answer

We apply the formula for arc length. Let $y'=\frac{\mathrm{d}y}{\mathrm{d}x}$, then we have 
$$
\sqrt{1+y'^2}=\frac{1}{6}\cdot\frac{3}{2}\cdot(x^2+4)^{1/2}\cdot2x=\frac{x}{2}\sqrt{x^2+4}
$$
Therefore, the length of the curve is given by 
$$
\int_0^4\sqrt{1+y'^2}\,\mathrm{d}x=\int_0^4\frac{x}{2}\sqrt{x^2+4}\,\mathrm{d}x=\frac{2}{3}\left(x^2+1\right)^{3/2}\Big|_{0}^{4}= \frac{26\sqrt{5}}{3}\approx 14.667.
$$
Thus the answer is (B).

Find the Length of the Curve y=16(x2+4)3/2,0≤x≤4y = \frac { 1 } { 6 } \left( x ^ { 2 } + 4 \right) ^ { 3 / 2 } , 0 \leq x \leq 4y=61​(x2+4)3/2,0≤x≤4

Find the Length of the Curve $y = \frac { 1 } { 6 } \left( x ^ { 2 } + 4 \right) ^ { 3 / 2 } , 0 \leq x \leq 4$