what is the integration of (e^-kt)
Answers
Answered by
7
Use substitution, letting u = -kt and du = -k dt:
∫e^(-kt) dt = ∫(e^u)(-du/k) = ∫(e^u)(-1/k)du
= (-1/k)∫e^u du [since k is a constant, we can pull that factor outside the integral]
= (-1/k)e^u + C [the integral of e^x dx is e^x]
Now reverse the earlier substitution:
= (-1/k)e^(-kt)+C
Which can also be written as:
-1/[ke^(kt)] + C
∫e^(-kt) dt = ∫(e^u)(-du/k) = ∫(e^u)(-1/k)du
= (-1/k)∫e^u du [since k is a constant, we can pull that factor outside the integral]
= (-1/k)e^u + C [the integral of e^x dx is e^x]
Now reverse the earlier substitution:
= (-1/k)e^(-kt)+C
Which can also be written as:
-1/[ke^(kt)] + C
Answered by
2
-1/k e^kt is its answer
Similar questions