How do you integrate #e^(-1.5x)#?

1 Answer
Jim H
Jun 27, 2015

Use substitution with #u = -1.5x#

Explanation:

Any integral of the form #int e^(kx) dx# where #k# is a constant, may be evaluated by using #u = kx#, so #du = k dx#

#int e^(kx) dx = 1/k int e^(kx) kdx#

# = 1/k int e^u du = 1/ke^u +C#

# = 1/ke^(kx) +C#

This answer makes sense, because the derivative of #e^(kx)# is #e^(kx) *k#
That is: we know the integral of #e^(-1.5x)# must involve #e^(-1.5x)#, but the derivative of that is #d/dx(e^(-1.5x))= -1.5 e^(-1.5x)#. We'll remove the #-1.5# by dividing:

#int e^(-1.5x) dx = 1/(-1.5) e^(-1.5x) +C#

Note: fraction with a decimal in the numerator or denominator looks strange to me. (I think: "Make up you mind! One or the other!")

#1.5 = 3/2# so #1/1.5 = 2/3# and we can write our answer:

#-2/3 e^(-1.5x) +C#

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