What is the integral of #e^(-0.2t) dt#?

1 Answer
Oct 19, 2015

Answer:

#int e^(-0.2t) dt = 5e^(-0.2t) +C#

Explanation:

Two methods:

Substitution

Let #u = 0.2t# so #du = -0.2 dt#

#int e^(-0.2t) dt# becomes

#-1/0.2 int e^u du = -1/(2/10) e^u +C#

# = -10/2 e^(-0.2t) +C#

# = -5 e^(-0.2t) +C#

Check the answer by differentiating.

Learn the rule

#int e^(ku) du = 1/k e^(ku) +C#

(Verify be differentiating.)

So

#int e^(-0.2t) dt = 1/(-0.2)e^(-0.2t) +C#

# = -5 e^(-0.2t) +C#