How can I show that the function F (x)= the integral from 2x to 5x of 1/ t dt is constant on the interval (0, +∞).
INT (1/t)dt=lnt over limits
= ln 5x -ln 2x= ln(5x/2x)=ln (2.5)
To show that the function F(x) = ∫(2x to 5x) (1/t) dt is constant on the interval (0, +∞), you can use the fundamental theorem of calculus.
Step 1: Calculate the derivative of F(x) with respect to x.
dF(x)/dx = d/dx ∫(2x to 5x) (1/t) dt
Step 2: Apply the fundamental theorem of calculus, which states that if F(x) is the integral of a function f(t) from a constant a to x, then F'(x) = f(x).
dF(x)/dx = (1/(5x)) * d(5x)/dx - (1/(2x)) * d(2x)/dx
= 1/5 - 1/2
= -1/10
Step 3: Notice that the derivative, dF(x)/dx, is a constant (-1/10).
Step 4: Since the derivative of F(x) is a constant, this indicates that F(x) must be a constant function. Therefore, F(x) = C, where C is any constant.
Step 5: Conclusion: The function F(x) is constant on the interval (0, +∞) because its derivative is always -1/10.
To show that the function F(x) is constant on the interval (0, +∞), we need to demonstrate that its derivative is zero over that entire interval.
First, let's find the derivative of F(x). 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 this case, F(x) is defined as the integral from 2x to 5x of 1/t dt. To find its derivative, we'll use the second part of the fundamental theorem of calculus and differentiate the integral limits and the integrand.
The derivative of F(x) with respect to x can be calculated using the chain rule:
F'(x) = d/dx ∫(2x to 5x) 1/t dt
Applying the chain rule, we differentiate the limits:
F'(x) = d/dx ∫(2x to 5x) 1/t dt = d/dx F(5x) - d/dx F(2x)
Next, we can apply the fundamental theorem of calculus. Since F(x) is defined as the integral of 1/t dt, we can evaluate F(5x) and F(2x) by anti-differentiating 1/t:
F'(x) = d/dx (ln|5x| - ln|2x|)
Now, we use the properties of logarithms and the chain rule to differentiate:
F'(x) = 1/x - 1/x = 0
Since the derivative of F(x) with respect to x is zero for all x in the interval (0, +∞), we can conclude that F(x) is constant over this interval.