How do you use substitution to integrate e−5xdx?

1 Answer
Jim H
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 ∫eudu.

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

∫e−5xdx=−15∫eudu=−15eu+C=−15e−5x+C

If you want to be different, turn it into ∫urdu

We know that e−5x=(ex)−5.

Alternative Method
If we want to make u=ex so that du=exdx, we need to change the exponent on (ex)−5. We write:

∫e−5xdx=∫(ex)−6xxdx.

Now with the substitution: u=ex so that du=exdx,

we get

∫e−5xdx=∫u−6du=u−5−5+C=−15(ex)−5+C

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

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