Show that the function F of x equals the integral from x to 3 times x of 1 over t dt is constant on the interval (0, +∞).

Question

Show that the function  is constant on the interval (0, +∞).

Answer 1

F(x) = ∫_x^3x1/tdt = lnt_x^3x = ln(3x) - ln(x) = ln3 = const.

We can prove this different way, by Fundamental Theorem of Calculus: F'(x) = 1/3x·3 - 1/x = 1/x - 1/x = 0

So F(x) = constant:

Answer 2

∫_x^3x(1/t)dt = ln(t) |_x^3x = ln|3x| - ln|x| = ln|3x/x| = ln(3) which is a constant