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For the function $ f(x) = \frac{1}{4} e^x + e^{-x} $, prove that the arc length on any interval has the same value as the area under the curve.

The arc length of the curve $y=f(x)$ on the interval $[a, b]$ is $L=\int_{a}^{b} \sqrt{1+\left[f^{\prime}(x)\right]^{2}} d x=\int_{a}^{b} \sqrt{[f(x)]^{2}} d x=\int_{a}^{b} f(x) d x,$ which is the area under the curve $y=f(x)$ on the interval $[a, b]$.

Applications of Integration

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it's Claressa when you read here. So the area under the curve it'll be from a to B 1/4 e to the X plus e to the negative x the X So we're first gonna find the derivative. When we get 1/4 e to the X minus e to the negative effects, we're gonna add one to the square of the derivative. So we get one plus won over 16. Eat the two x minus 1/4 B to the X E to the negative X when it's 1/4 Ito EXT. E to the negative X plus eats and negative two acts, which is equal to one plus one over 16 e to the two X minus one. Huh? Less e to the negative to X, which is equal to one over 16 8 the two x plus 1/2. So let's e to the negative to X, And when we factor this cell comes one over four feet X plus e to the negative X square. So we're just gonna put this into our arguing formula. So from a to B square, root well 1/4 e to the X plus eats and negative X square D X, which is equal to a to B when fourth, Eat the axe wa seat the negative x t x and this is equal to the area under the curse.