Find the average value of f(x) = 1 over x over the interval [e, 2e].
the natural logarithm of 2 over e
negative 1 over 2 times e squared
Ln2 <- my anser
Ln3
avg value is ∫[a,b]f(x)/(b-a)
= lnx[e,2e] / (2e-e)
= (ln(2e)-lne)/e
= (ln2+lne-lne)/e
= ln2/e
you forgot to divide by the interval width
To find the average value of a function over an interval, you need to integrate the function over that interval and then divide the result by the length of the interval. The average value of f(x) = 1/x over the interval [e, 2e] can be calculated as follows:
1. Integrate the function f(x) = 1/x with respect to x over the interval [e, 2e]:
∫(1/x) dx = ln|x| + C,
where C is the constant of integration.
2. Evaluate the definite integral over the interval [e, 2e]:
(∫(1/x) dx)[e,2e] = ln|2e| − ln|e| = ln(2e) − ln(e) = ln(2).
3. Calculate the length of the interval [e, 2e]:
Length = 2e - e = e.
4. Divide the result from step 2 by the length from step 3:
Average value = ln(2) / e.
So, the average value of f(x) = 1/x over the interval [e, 2e] is ln(2) / e.
To find the average value of the function f(x) = 1/x over the interval [e, 2e], follow these steps:
1. Start by calculating the definite integral of f(x) over the given interval. In this case, we have:
∫[e, 2e] (1/x) dx
2. Evaluate the integral. The integral of 1/x is ln|x|, so we have:
ln|x| | from e to 2e
3. Plug in the upper and lower limits of integration:
ln|2e| - ln|e|
4. Simplify the expression:
ln(2e) - ln(e) = ln(2) + ln(e) - ln(e) = ln(2)
5. Lastly, divide the result by the length of the interval. The length of the interval [e, 2e] is 2e - e = e, so we divide ln(2) by e:
ln(2) / e
Therefore, the average value of f(x) = 1/x over the interval [e, 2e] is ln(2) / e.