Sure, I'd be happy to help with this integral function problem. First, let's start by defining the function F(x) as the integral of 1/x from 1 to x. This can be written as:
F(x) = ∫(1/x)dx from 1 to x
Now, we can use the fundamental theorem of calculus to solve for F(x). Since the derivative of F(x) is 1/x, we can integrate 1/x to get back to F(x). This means that:
F'(x) = 1/x = F(x)
Next, we can use the initial conditions given in the problem to solve for the constant of integration. For part (a), we know that F(2) = 0, so we can plug in 2 for x and set it equal to 0:
F(2) = ∫(1/x)dx from 1 to 2 = 0
Solving this integral, we get:
ln(2) - ln(1) = 0
ln(2) = 0
This means that ln(2) must equal 0, which is not true. Therefore, there is no solution for F(x) that satisfies the initial condition F(2) = 0.
For part (b), we know that F(2) = -3, so we can plug in 2 for x and set it equal to -3:
F(2) = ∫(1/x)dx from 1 to 2 = -3
Solving this integral, we get:
ln(2) - ln(1) = -3
ln(2) = -3 + ln(1)
ln(2) = -3
Therefore, the function F(x) = ∫(1/x)dx from 1 to x with the initial condition F(2) = -3 satisfies the given conditions. I hope this helps you with your integral function problem. Let me know if you have any further questions or need clarification. Good luck!
