The average (mean) value of (1/x) over the interval 1<= x<= e is
a)1
b)(1/e)
c)(1/e^2)-1
d)(1+e)/2
e) 1/(e-1)
(ln e - ln1)/(e-1) = 1/(e-1)
Well, it seems like you're asking for the average value of 1/x over the interval 1 <= x <= e. Let's think about this for a moment...
First, we need to find the definite integral of 1/x from 1 to e. Integrating 1/x gives us ln(x), so integrating ln(x) from 1 to e should give us the answer.
Let's evaluate that integral: ∫[1 to e] (1/x) dx = ln(x) [1 to e] = ln(e) - ln(1) = 1 - 0 = 1.
So, the definite integral of 1/x over the interval 1 <= x <= e is equal to 1.
Now, to find the average value, we divide the definite integral by the length of the interval. The length of the interval is e - 1.
Therefore, the average value of 1/x over the interval 1 <= x <= e is 1 / (e - 1).
So, the option e) 1/(e - 1) seems to be the correct answer.
To find the average value of the function (1/x) over the interval 1 <= x <= e, we need to find the definite integral of (1/x) over that interval and divide it by the length of the interval.
The definite integral of (1/x) with respect to x over the interval 1 to e is:
∫(1/x) dx from 1 to e
This integral can be evaluated as the natural logarithm of e minus the natural logarithm of 1:
ln(e) - ln(1)
Since ln(e) is equal to 1 and ln(1) is equal to 0, we have:
1 - 0
Therefore, the definite integral of (1/x) with respect to x over the interval 1 to e is equal to 1.
The length of the interval 1 to e is e - 1.
To find the average value, we divide the definite integral by the length of the interval:
1 / (e - 1)
So, the average value of (1/x) over the interval 1 <= x <= e is 1 / (e - 1).
Therefore, the correct answer is e) 1/(e-1).
To find the average value of a function over an interval, you need to compute the definite integral of the function over that interval and then divide it by the length of the interval. In this case, we want to find the average value of the function f(x) = 1/x over the interval 1 <= x <= e.
First, let's calculate the definite integral of f(x) = 1/x over the interval [1, e]. We can do this by integrating the function:
∫(1/x) dx
Using the rule of integration for ln(x), we have:
= ln(x) + C
Evaluating the definite integral over the interval [1, e]:
∫(1/x) dx = [ln(x)] with limits 1 to e
= ln(e) - ln(1)
= 1 - 0
= 1
Now, we need to divide the integral by the length of the interval, which is e - 1:
Average value = (1/(e - 1)) * ∫(1/x) dx
Substituting the integral value we found earlier:
Average value = (1/(e - 1)) * 1
= 1/(e - 1)
Therefore, the correct answer is e) 1/(e - 1). The average value of (1/x) over the interval 1 <= x <= e is 1/(e - 1).