Integrals and Arc Length, Calculus Help:

Show that the function f(x) = 1/2 (eˆx + eˆ−x)
has the following curious property: On any
interval [a, b], the arc length L of the graph
of f is equal to the area A under the graph.

Note that (e^x + e^(-x))/2 = cosh(x). Then, ∫cosh(x) x=a..b = sinh(x) Then, find arc length by

s = ∫(√(1+f'(x)^2)dx
= ∫(√(1+sinh(x)^2)dx

Then from the identity
cosh(x)^2 - sinh(x)^2 = 1
sinh(x)^2 +1 = cosh(x)^2

So,

s = ∫(√(1+sinh(x)^2)dx
s = ∫(√(cosh(x)^2)dx
s = ∫(cosh(x)dx

Now note that

A = ∫(cosh(x)dx

So, arc length must be equal to area for any interval (a,b).

Hope that helps!

There aren't really criteria (that I know of) for arbitrary integrals to be equal, but we don't really need that in this case. We have the same interval and something curious will happen when these two things are evaluated. The area under the curve is just
∫[a,b] f(x) dx

but the arc length is
∫[a,b] √(1 + f'(x)²) dx

Now if it so happens that
√(1 + f'(x)²) = f(x)

on the interval [a,b], then those two integrals are obviously equal as they're identical. That's what happens here. Now just do the algebra to show that.

Arc length S = integral(x=a to b)(sqrt(1 + [f'(x)]^2) dx) =

integral(x=a to b)(sqrt(1 + (1/4)*(e^x - e^(-x))^2) dx) =

integral(x=a to b)((1/2)*sqrt(4 + e^(2x) + e^(-2x) - 2) dx) =

integral(x=a to b)((1/2)*sqrt(e^(2x) + e^(-2x) + 2) dx) =

integral(x= a to b)((1/2)*sqrt((e^x + e^-x)^2) dx) =

integral(x=a to b)((1/2)*(e^x + e^(-x)) dx) =

integral(x=a to b)(f(x) dx) = area under f(x) = (1/2)*(e^x + e^(-x)) on [a,b].

Note that sqrt((e^x + e^(-x))^2) = e^x + e^(-x) since (e^x + e^(-x)) > 0 for all x.