How do you use substitution to integrate #e^ (-5x) dx#?

1 Answer
May 25, 2015

Here are two solutions:

The "usual way" (as far as my experience), is to turn this into an integral of the form #int e^u du#.

Let #u=-5x#, this makes #du = -5 dx#, so the integral becomes:

#int e^(-5x) dx = -1/5 int e^u du = -1/5 e^u +C = -1/5 e^(-5x) +C#

If you want to be different, turn it into #int u^r du#

We know that #e^(-5x) = (e^x)^-5#.

Alternative Method
If we want to make #u= e^x# so that #du = e^x dx#, we need to change the exponent on #(e^x)^-5#. We write:

#int e^(-5x) dx = int (e^x)^-6 x^x dx#.

Now with the substitution: #u= e^x# so that #du = e^x dx#,

we get

#int e^(-5x) dx = int u^-6 du = (u^(-5))/-5 +C = -1/5 (e^x)^-5 +C#

Which is equal to the answer by the more usual method.