set up and evaluate the integral to find the exact length of the curve.\ng(y)=e^{y}+\frac{1}{4} e^{-y}, 0…

Explanation:

Step1: Find the derivative of (g(y))

The derivative of (g(y)=e^{y}+\frac{1}{4}e^{-y}) with respect to (y) is (g^{\prime}(y)=e^{y}-\frac{1}{4}e^{-y}) (using the rule (\frac{d}{dy}(e^{ay}) = ae^{ay})).

Step2: Use the arc - length formula for a function (x = g(y))

The arc - length formula for (x = g(y)) on the interval ([a,b]) is (L=\int_{a}^{b}\sqrt{1+(g^{\prime}(y))^{2}}dy). First, find ((g^{\prime}(y))^{2}=(e^{y}-\frac{1}{4}e^{-y})^{2}=e^{2y}-\frac{1}{2}+\frac{1}{16}e^{-2y}). Then (1+(g^{\prime}(y))^{2}=1 + e^{2y}-\frac{1}{2}+\frac{1}{16}e^{-2y}=e^{2y}+\frac{1}{2}+\frac{1}{16}e^{-2y}=(e^{y}+\frac{1}{4}e^{-y})^{2}) (using the formula ((a + b)^{2}=a^{2}+2ab + b^{2}) where (a = e^{y}) and (b=\frac{1}{4}e^{-y})). So, (\sqrt{1+(g^{\prime}(y))^{2}}=e^{y}+\frac{1}{4}e^{-y}).

Step3: Evaluate the integral

Now, evaluate the integral (L=\int_{0}^{8}(e^{y}+\frac{1}{4}e^{-y})dy). Using the integral rules (\int e^{ay}dy=\frac{1}{a}e^{ay}+C) ((a\neq0)), we have: (\int_{0}^{8}(e^{y}+\frac{1}{4}e^{-y})dy=\left[e^{y}-\frac{1}{4}e^{-y}\right]_{0}^{8}) [=(e^{8}-\frac{1}{4}e^{-8})-(e^{0}-\frac{1}{4}e^{0})] [=e^{8}-\frac{1}{4}e^{-8}-(1 - \frac{1}{4})] [=e^{8}-\frac{1}{4}e^{-8}-\frac{3}{4}]

Answer:

(e^{8}-\frac{1}{4}e^{-8}-\frac{3}{4})