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

Take an arbitrary interval [a, b], where a < b.

Compute the arc length L of y = f(x) over [a, b] :

[tex]L=\displaystyle\int_a^b\sqrt{1+\left(f'(x)\right)^2}\,\mathrm dx[/tex]

Now comptue the area A under the curve y = f(x) over [a, b] :

[tex]A=\displaystyle\int_a^bf(x)\,\mathrm dx[/tex]

We have

f (x) = 1/4 e ⁻ˣ + e ˣ   →   f ' (x) = -1/4 e ⁻ˣ + e ˣ

Then

√(1 + (f ' (x))²) = √(1 + (-1/4 e ⁻ˣ + e ˣ)²)

… = √(1 + 1/16 e ⁻²ˣ - 1/2 + e ²ˣ)

… = √(1/16 e ⁻²ˣ + 1/2 + e ²ˣ)

… = 1/4 √(e ⁻²ˣ + 8 + 16e ²ˣ)

… = 1/4 √((e ⁻ˣ + 4 e ˣ)²)

… = 1/4 (e ⁻ˣ + 4 e ˣ)

… = 1/4 e ⁻ˣ + e ˣ

… = f (x)

so both A = L for any choice of interval [a, b].

It is true that the arc length on any interval has the same value as the area under the curve.

The function is given as:

[tex]f(x) = \frac 14e^{-x} + e^x[/tex]

Differentiate the function

[tex]f'(x) = -\frac 14e^{-x} + e^x[/tex]

On any interval, the following must be true

[tex]f(x) =f'(x)[/tex]

and

[tex]f(x) = \sqrt{1 + (f'(x))^2}[/tex]

So, we have:

[tex]f(x) = \sqrt{1 + (-\frac 14e^{-x} + e^x)^2}[/tex]

Expand the exponents

[tex]f(x) = \sqrt{1 + (\frac{1}{16}e^{-2x} - \frac 12 + e^{2x})}[/tex]

Remove the bracket

[tex]f(x) = \sqrt{1 + \frac{1}{16}e^{-2x} - \frac 12 + e^{2x}}[/tex]

Evaluate the like terms

[tex]f(x) = \sqrt{\frac{1}{16}e^{-2x} + \frac 12 + e^{2x}}[/tex]

Multiply by 16/16

[tex]f(x) = \sqrt{\frac{16}{16}(\frac{1}{16}e^{-2x} + \frac 12 + e^{2x})}[/tex]

So, we have:

[tex]f(x) = \sqrt{\frac{1}{16}(e^{-2x} + 8 + 16e^{2x})}[/tex]

Take the square root of 1/16

[tex]f(x) = \frac{1}{4}\sqrt{e^{-2x} + 8 + 16e^{2x}}[/tex]

Express the radical as a perfect square

[tex]f(x) = \frac{1}{4}\sqrt{(e^{-x} + 4e^{x})^2}[/tex]

Evaluate the exponents

[tex]f(x) = \frac{1}{4} * (e^{-x} + 4e^{x})[/tex]

Evaluate the products

[tex]f(x) = \frac{1}{4}e^{-x} + e^{x}[/tex]

Hence, it has been proved that  the arc length on any interval has the same value as the area under the curve.

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