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 $\frac{d}{dx}(\frac{u^{3}}{3})=u^{2}=(x + 1)^{2}$ and $\frac{d}{dx}(\frac{1}{4u})=-\frac{1}{4u^{2}}=-\frac{1}{4(x + 1)^{2}}$. So, $y'=(x + 1)^{2}-\frac{1}{4(x + 1)^{2}}$.

Step2: Use the arc - length formula

The arc - length formula for a curve $y = f(x)$ from $x=a$ to $x=b$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. Calculate $1+(y')^{2}$: [ \begin{align*} 1+(y')^{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 $\sqrt{1+(y')^{2}}=(x + 1)^{2}+\frac{1}{4(x + 1)^{2}}$.

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}\ &=(\frac{1}{3}+1 + 1-\frac{1}{8})-(0 + 0+0-\frac{1}{4})\ &=\frac{8 + 24+24 - 3}{24}+\frac{1}{4}\ &=\frac{53}{24}+\frac{6}{24}\ &=\frac{59}{24} \end{align*} ]

Answer:

$\frac{59}{24}$