Answer
The arc length on any interval has the same value as the area under the curve.
Work Step by Step
$f(x)=\frac{1}{4}e^x+e^{-x}$
$\int_a^b\frac{1}{4}e^x+e^{-x}dx=\Big[\frac{e^x}{4}-e^{-x}\Big]_a^b$$=\frac{e^b}{4}-e^{-b}-\frac{e^a}{4}+e^{-a}$
$\frac{dy}{dx}=\frac{e^x}{4}-e^{-x}$
$L=\int_a^b\sqrt{1+\Big(\frac{dy}{dx}\Big)^2}dx=\Big[\frac{e^{-x}(e^{2x}-4)}{4}\Big]_a^b$
$=\frac{e^{-b}(e^{2b}-4)-e^{-a}(e^{2a}-4)}{4}$
$=\frac{e^b}{4}-e^{-b}-\frac{e^a}{4}+e^{-a}=\int_a^b\frac{1}{4}e^x+e^{-x}dx$
