Question

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.

Answer 1

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

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

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

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

$A=\displaystyle\int_a^bf(x)\,\mathrm dx$

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].