a)Find the average value of f(x) = e-x over the interval [0,k]
b)What happens to the average value of f on [0,k] as k approaches infinity
a) To find the average value of a function f(x) over an interval [a, b], we need to calculate the definite integral of the function over that interval and then divide it by the width of the interval (b - a).
For the function f(x) = e^(-x) and the interval [0, k], we can find the average value as follows:
1. Calculate the integral of f(x) over the interval [0, k]:
∫(0 to k) e^(-x) dx
To evaluate this integral, we can use the property of the exponential function: ∫e^(-x) dx = -e^(-x) + C
Thus, evaluating the definite integral, we get:
[-e^(-x)] evaluated from 0 to k
= -e^(-k) - (-e^(-0))
= -e^(-k) + 1
2. Divide the result by the width of the interval ([0, k]):
average value = [∫(0 to k) e^(-x) dx] / (k - 0)
= (-e^(-k) + 1) / k
So the average value of f(x) = e^(-x) over the interval [0, k] is (-e^(-k) + 1) / k.
b) To determine what happens to the average value of f on [0, k] as k approaches infinity, we can take the limit of the expression we derived in part a) as k approaches infinity.
Lim(k->∞) [(-e^(-k) + 1) / k]
We can observe that as k approaches infinity, the term e^(-k) approaches 0, as the exponent becomes increasingly negative. Therefore, the average value of f approaches:
Lim(k->∞) [(1) / k]
= 0
Hence, as k approaches infinity, the average value of f(x) = e^(-x) on [0, k] becomes zero.