Find the arc length of the graph of the given equation on the specified interval.
$$y = \frac { 2 } { 3 } \left( x ^ { 2 } + 1 \right) ^ { 3 / 2 } , [ 2,3 ]$$
A) $\frac { 41 } { 3 }$
B) $\frac { 41 } { 2 }$
C) $\frac { 41 } { 4 }$
D) $\frac { 41 } { 6 }$

Question

Quizplus · Accepted Answer

The arc length $L$ of a curve defined by $y = f(x)$ from $x = a$ to $x = b$ is given by the integral $$L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx.$$ For $y = \frac{2}{3}(x^2 + 1)^{3/2}$, we first find $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2}(x^2 + 1)^{1/2} \cdot 2x = 2x(x^2 + 1)^{1/2}.$$ Then, $\left(\frac{dy}{dx}\right)^2 = 4x^2(x^2 + 1)$. Substituting into the arc length formula gives: $$L = \int_{2}^{3} \sqrt{1 + 4x^2(x^2 + 1)} dx = \int_{2}^{3} \sqrt{1 + 4x^2 + 4x^4} dx.$$ Simplifying inside the square root yields: $$L = \int_{2}^{3} \sqrt{4x^4 + 4x^2 + 1} dx = \int_{2}^{3} \sqrt{(2x^2 + 1)^2} dx = \int_{2}^{3} (2x^2 + 1) dx.$$ Integrating gives: $$L = \left[\frac{2}{3}x^3 + x\right]_{2}^{3} = \left(\frac{2}{3}(27) + 3\right) - \left(\frac{2}{3}(8) + 2\right) = \frac{54}{3} + 3 - \frac{16}{3} - 2 = 18 + 3 - \frac{16}{3} - 2 = \frac{41}{3}.$$

Find the Arc Length of the Graph of the Given y=23(x2+1)3/2,[2,3]y = \frac { 2 } { 3 } \left( x ^ { 2 } + 1 \right) ^ { 3 / 2 } , [ 2,3 ]y=32​(x2+1)3/2,[2,3]

Find the Arc Length of the Graph of the Given $y = \frac { 2 } { 3 } \left( x ^ { 2 } + 1 \right) ^ { 3 / 2 } , [ 2,3 ]$