a barn is 100 feet long and 40 feet wide (see figure). a cross - section of the roof is the inverted…

Explanation:

Step1: Recall arc - length formula for a curve

The arc - length formula for a curve $y = f(x)$ from $x=a$ to $x = b$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. First, find the derivative of $y = 31-10(e^{x/20}+e^{-x/20})$. Using the chain - rule, if $y = 31-10e^{x/20}-10e^{-x/20}$, then $y'=-10\times\frac{1}{20}e^{x/20}+10\times\frac{1}{20}e^{-x/20}=-\frac{1}{2}(e^{x/20}-e^{-x/20})$.

Step2: Calculate $(y')^{2}$

$(y')^{2}=\frac{1}{4}(e^{x/10}-2 + e^{-x/10})$.

Step3: Calculate $1+(y')^{2}$

$1+(y')^{2}=1+\frac{1}{4}(e^{x/10}-2 + e^{-x/10})=\frac{1}{4}(e^{x/10}+2 + e^{-x/10})=\left[\frac{1}{2}(e^{x/20}+e^{-x/20})\right]^{2}$.

Step4: Calculate $\sqrt{1+(y')^{2}}$

$\sqrt{1+(y')^{2}}=\frac{1}{2}(e^{x/20}+e^{-x/20})$.

Step5: Calculate the arc - length of the cross - section

The cross - section of the roof is from $x=-20$ to $x = 20$. The arc - length $L$ of the cross - section is $L=\int_{-20}^{20}\frac{1}{2}(e^{x/20}+e^{-x/20})dx$. Since $\int\frac{1}{2}(e^{x/20}+e^{-x/20})dx = 10(e^{x/20}-e^{-x/20})+C$, then $L=\left[10(e^{x/20}-e^{-x/20})\right]_{-20}^{20}=10(e - e^{-1})-10(e^{-1}-e)=20(e - e^{-1})$.

Step6: Calculate the surface area of the roofing

The length of the barn is $l = 100$ feet. The surface area $A$ of the roofing is $A = 100\times L$. $A = 100\times20(e - e^{-1})=2000(e - e^{-1})\approx2000(2.71828 - 0.36788)=2000\times2.3504\approx4701$.

Answer:

$4701$