3. (25 points) find the arc length of the curve (y = \frac{(x + 1)^3}{3}+\frac{1}{4x + 4}) between (x = 0)…

Explanation:

Step1: Find the derivative of (y)

First, rewrite (y=\frac{(x + 1)^{3}}{3}+\frac{1}{4(x + 1)}). Using the power - rule, if (u=x + 1), then (y=\frac{u^{3}}{3}+\frac{1}{4u}). (y^\prime=(x + 1)^{2}-\frac{1}{4(x + 1)^{2}})

Step2: Use the arc - length formula

The arc - length formula for (y = f(x)) from (x=a) to (x=b) is (L=\int_{a}^{b}\sqrt{1+(y^\prime)^{2}}dx). First, find ((y^\prime)^{2}=(x + 1)^{4}-\frac{1}{2}+\frac{1}{16(x + 1)^{4}}). Then (1+(y^\prime)^{2}=(x + 1)^{4}+\frac{1}{2}+\frac{1}{16(x + 1)^{4}}=\left((x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right)^{2}). So, (\sqrt{1+(y^\prime)^{2}}=(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}).

Step3: Calculate the definite integral

(L=\int_{0}^{1}\left((x + 1)^{2}+\frac{1}{4(x + 1)^{2}}\right)dx). Let (u=x + 1), (du=dx). When (x = 0), (u = 1); when (x = 1), (u = 2). (L=\int_{1}^{2}\left(u^{2}+\frac{1}{4u^{2}}\right)du=\int_{1}^{2}\left(u^{2}+\frac{1}{4}u^{-2}\right)du). Integrating term - by - term: (\int\left(u^{2}+\frac{1}{4}u^{-2}\right)du=\frac{u^{3}}{3}-\frac{1}{4u}+C). Evaluating the definite integral: (\left[\frac{u^{3}}{3}-\frac{1}{4u}\right]_{1}^{2}=\left(\frac{2^{3}}{3}-\frac{1}{4\times2}\right)-\left(\frac{1^{3}}{3}-\frac{1}{4\times1}\right)) (=\left(\frac{8}{3}-\frac{1}{8}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{8}{3}-\frac{1}{8}-\frac{1}{3}+\frac{1}{4}) (=\left(\frac{8}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)=\frac{7}{3}+\frac{1}{8}=\frac{56 + 3}{24}=\frac{59}{24})

Answer:

(\frac{59}{24})