How do I integrate 1-e^-kt from 0 to infinity?
What is the derivative of t+1/k e^-kt
which means.....when t=inf, the integral is meaningless.
To integrate the function 1-e^(-kt) from 0 to infinity, follow these steps:
Step 1: Set up the integral.
∫(0 to ∞) (1 - e^(-kt)) dt
Step 2: Apply the integral limits.
To evaluate this definite integral, we will determine the limit as the upper bound approaches infinity.
Step 3: Integrate the function.
To integrate 1 - e^(-kt), we need to find the antiderivative of e^(-kt).
The antiderivative of e^(-kt) with respect to t is -1/k * e^(-kt).
So, the integral becomes:
∫(0 to ∞) (1 - e^(-kt)) dt = [t + (1/k) * e^(-kt)] evaluated from 0 to ∞
Step 4: Evaluate the integral limits.
Now, let's plug in the upper and lower bounds to find the value of the integral:
[t + (1/k) * e^(-kt)] evaluated from 0 to ∞
When the upper bound is infinity, we have:
lim as t → ∞ [(t + (1/k) * e^(-kt))] - [(0 + (1/k) * e^(-k * 0))]
Simplifying further, we have:
lim as t → ∞ [(t + (1/k) * e^(-kt))] - 0
Taking the limit as t approaches infinity, we get:
lim as t → ∞ [(t + 0)] = ∞
So, the integral of 1 - e^(-kt) from 0 to infinity is equal to infinity.