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

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 (y^\prime=(x + 1)^{2}-\frac{1}{4(x + 1)^{2}}).

Step2: Use the arc - length formula

The arc - length formula for a function (y = f(x)) from (x=a) to (x=b) is (L=\int_{a}^{b}\sqrt{1+(y^\prime)^{2}}dx). First, find (1+(y^\prime)^{2}): [ \begin{align*} 1+(y^\prime)^{2}&=1+((x + 1)^{2}-\frac{1}{4(x + 1)^{2}})^{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}}\ &=((x + 1)^{2}+\frac{1}{4(x + 1)^{2}})^{2} \end{align*} ] Then (\sqrt{1+(y^\prime)^{2}}=(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}).

Step3: Calculate the definite integral

[ \begin{align*} L&=\int_{0}^{1}((x + 1)^{2}+\frac{1}{4(x + 1)^{2}})dx\ &=\int_{0}^{1}(x^{2}+2x + 1+\frac{1}{4}(x + 1)^{-2})dx\ &=\left[\frac{x^{3}}{3}+x^{2}+x-\frac{1}{4(x + 1)}\right]_{0}^{1}\ &=\left(\frac{1}{3}+1 + 1-\frac{1}{8}\right)-\left(0 + 0+0-\frac{1}{4}\right)\ &=\frac{8 + 24+24 - 3}{24}+\frac{1}{4}\ &=\frac{53}{24}+\frac{6}{24}\ &=\frac{59}{24} \end{align*} ]

Answer:

(\frac{59}{24})