Show that the function F(x) = integral[x to 3x](1/t)dt is constant on the interval (0, +∞).
To show that the function F(x) = ∫[x to 3x] (1/t) dt is constant on the interval (0, +∞), we need to prove that its derivative with respect to x is zero.
Let's begin by finding the derivative of F(x) with respect to x using the Fundamental Theorem of Calculus.
The Fundamental Theorem of Calculus states that if F(x) is defined as the integral of a function f(t), then the derivative of F(x) with respect to x is equal to f(x). In other words, if F(x) = ∫[a to x] f(t) dt, then dF(x)/dx = f(x).
In our case, F(x) = ∫[x to 3x] (1/t) dt. To find dF(x)/dx, we differentiate both sides of the equation with respect to x:
d/dx [F(x)] = d/dx ∫[x to 3x] (1/t) dt.
Now, to find the derivative of F(x), we'll use the Leibniz rule for differentiating definite integrals with variable limits of integration. According to the rule, if F(x) = ∫[a(x) to b(x)] f(t) dt, the derivative of F(x) with respect to x is:
dF(x)/dx = f(b(x))*db(x)/dx - f(a(x))*da(x)/dx.
In our case, a(x) = x and b(x) = 3x. Therefore, we have:
d/dx [F(x)] = (1/3x)*(d(3x)/dx) - (1/x)*(d(x)/dx).
Simplifying further, we have:
d/dx [F(x)] = (1/3x)*(3) - (1/x)*(1).
d/dx [F(x)] = 1/x - 1/x.
Notice that the two terms on the right side of the equation cancel each other out, resulting in:
d/dx [F(x)] = 0.
Since the derivative of F(x) with respect to x is zero, this implies that F(x) is constant on the interval (0, +∞).
Therefore, we have shown that the function F(x) = ∫[x to 3x] (1/t) dt is constant on the interval (0, +∞).
F' = (1/(3x))*3 - (1/x)*1 = 1-1 = 0
so, F is constant