3. (25 points) find the arc length of the curve y = ((x + 1)^3)/3 + 1/(4x + 4) between x = 0 and x = 1…

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: Calculate (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*} ]

Step3: Use the arc - length formula (L=\int_{a}^{b}\sqrt{1+(y^\prime)^{2}}dx)

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

Step4: Evaluate the definite integral

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

Answer:

(\frac{59}{24})