3. (25 points) find the arc length of the curve\n\ny = \frac{(x + 1)^3}{3}+\frac{1}{4x + 4}\n\nbetween (x =…

Explanation:

Step1: Find the derivative of (y)

First, rewrite (y=\frac{(x + 1)^{3}}{3}+\frac{1}{4(x + 1)}). Using the power - rule ((u^{n})^\prime=nu^{n - 1}u^\prime), the derivative of (\frac{(x + 1)^{3}}{3}) is ((x + 1)^{2}), and the derivative of (\frac{1}{4(x + 1)}=\frac{1}{4}(x + 1)^{-1}) is (-\frac{1}{4}(x + 1)^{-2}). So (y^\prime=(x + 1)^{2}-\frac{1}{4(x + 1)^{2}}).

Step2: Calculate (1+(y^\prime)^{2})

[ \begin{align*} 1+(y^\prime)^{2}&=1+\left[(x + 1)^{2}-\frac{1}{4(x + 1)^{2}}\right]^{2}\ &=1+(x + 1)^{4}-\frac{1}{2}+\frac{1}{16(x + 1)^{4}}\ &=(x + 1)^{4}+\frac{1}{2}+\frac{1}{16(x + 1)^{4}}\ &=\left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]^{2} \end{align*} ]

Step3: Use the arc - length formula

The arc - length formula is (L=\int_{a}^{b}\sqrt{1+(y^\prime)^{2}}dx), where (a = 0), (b = 1). Since (\sqrt{1+(y^\prime)^{2}}=(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}), then (L=\int_{0}^{1}\left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]dx).

Step4: Integrate term - by - term

[ \begin{align*} \int_{0}^{1}\left[(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right]dx&=\int_{0}^{1}(x + 1)^{2}dx+\frac{1}{4}\int_{0}^{1}(x + 1)^{-2}dx\ &=\left[\frac{(x + 1)^{3}}{3}\right]{0}^{1}+\frac{1}{4}\left[-\frac{1}{x + 1}\right]{0}^{1}\ &=\left(\frac{(1 + 1)^{3}}{3}-\frac{(0 + 1)^{3}}{3}\right)+\frac{1}{4}\left(-\frac{1}{1+1}+\frac{1}{0 + 1}\right)\ &=\left(\frac{8}{3}-\frac{1}{3}\right)+\frac{1}{4}\left(-\frac{1}{2}+1\right)\ &=\frac{7}{3}+\frac{1}{8}\ &=\frac{56 + 3}{24}\ &=\frac{59}{24} \end{align*} ]

Answer:

(\frac{59}{24})