Show that the function F( x)=∫(from x to 3x)1/t dt is constant on the interval (0, +∞).
Okay so I'm not sure if I should use the FTOC Part 1 or something else but this question is just really confusing.
I asked my brother and he said I should just solve the integral but all I keep getting is:
F(x) = ln|3x| - ln|x|
It feels as though I'm forgetting the chain rule or something but I don't know how to implement it.
Even if I wanted to find F'(x) or something else, the question itself doesn't make sense to me. What is it asking for? Thank you.
ln(3x)-ln(x) = ln(3x/x) = ln3
is constant
To show that the function F(x) = ∫(from x to 3x)1/t dt is constant on the interval (0, +∞), we need to differentiate F(x) and prove that the derivative is equal to zero.
Let's start by finding the derivative of F(x) using the Fundamental Theorem of Calculus, Part 1 (FTC 1). According to FTC 1, if a function F(x) is defined as the integral of another function f(t) with respect to t from a constant to x, then the derivative of F(x) is f(x).
So, in this case, using FTC 1, we can write:
F'(x) = d/dx (∫(from x to 3x)1/t dt)
Now, we need to evaluate the derivative of the integral with respect to x. To do this, we can use the chain rule. The chain rule states that d/dx (f(g(x))) = f'(g(x)) * g'(x).
Let's take a closer look at the integral:
∫(from x to 3x)1/t dt
The function inside the integral is 1/t. To apply the chain rule, let's define a new function u(x) = 3x. Then, the integral becomes:
∫(from x to u(x))1/t dt
Now, we can rewrite the derivative F'(x) using the chain rule:
F'(x) = d/du (∫(from x to u(x))1/t dt) * du/dx
The second part, du/dx, is simply the derivative of 3x, which is 3.
Now, we need to find the derivative of ∫(from x to u(x))1/t dt with respect to u. This is where the rule for differentiating integrals with variable limits comes in.
Let's define a new function G(u) as:
G(u) = ∫(from x to u)1/t dt
Now, we want to differentiate G(u) with respect to u:
d/dx (∫(from x to u(x))1/t dt) = d/dx(G(u))
Using the chain rule again:
d/dx(G(u)) = dG(u)/du * du/dx
Now, we can combine the two derivatives we have:
F'(x) = d/du (∫(from x to u(x))1/t dt) * du/dx
F'(x) = d/dx(G(u)) * du/dx
F'(x) = dG(u)/du * du/dx
Notice that dG(u)/du represents the derivative of G(u) and does not depend on x. Therefore, it is a constant.
Since du/dx = 3, the whole expression F'(x) simplifies to:
F'(x) = Constant * 3
F'(x) = 3 * Constant
Now, since F'(x) is a constant (independent of x), we can conclude that the original function F(x) is constant on the interval (0, +∞).
To show that the function F(x) is constant on the interval (0, +∞), we need to prove that F'(x) = 0 for all x in the given interval. This means taking the derivative of the function F(x) with respect to x and showing that it results in zero.
First, let's calculate F(x) using the given integral:
F(x) = ∫(from x to 3x) 1/t dt
To evaluate this integral, we can use the properties of logarithms. Recall that for any positive constant a, the integral of 1/t with respect to t from a to b can be expressed as ln|b| - ln|a|.
Applying this rule to our integral, we have:
F(x) = ln|3x| - ln|x|
Now, let's find the derivative of F(x) using the chain rule:
F'(x) = [d/dx (ln|3x|)] - [d/dx (ln|x|)]
To differentiate ln|3x| and ln|x|, we can use the rule for the derivative of logarithmic functions. The derivative of ln|u| with respect to u is 1/u times the derivative of u with respect to x.
Differentiating ln|3x| gives us:
[d/dx (ln|3x|)] = (1/(3x))(d/dx (3x)) = 1/x
Differentiating ln|x| gives us:
[d/dx (ln|x|)] = (1/x)(d/dx (x)) = 1/x
Now let's substitute these results back into the expression for F'(x):
F'(x) = 1/x - 1/x = 0
Notice that the derivative of F(x) is zero, F'(x) = 0, which implies that F(x) is constant on the interval (0, +∞).
Therefore, the function F(x) = ∫(from x to 3x) 1/t dt is indeed constant on the interval (0, +∞).