Question #942b1

1 Answer
May 23, 2015

Mike, all three can be done by substitution. Let #u# (or #g(x)# depending on the notation you've been taught) be the exponent on #e#.

I'll go through the first one, {EDIT: I'll use notation and an approach that may be more familiar to you

#int_0^3 e^(-3x) dx#

First let's just find the indefinite integral:

#int e^(-3x) dx#

Let #u=-3x#,

That makes #du = -3 dx#.

I don't see a #-3# in front of the #dx# in the integrand, so We'll solve for #dx#.

#dx=-1/3 du#, now here is what we have:

#int e^(overbracecolor(red)(-3x)^(color(red)"u")) overbrace color(green)(dx)^color(green)(-1/3 du) = int e^color(red)(u) * color(green)((-1/3) du) = -1/3 int e^u du = -1/3 e^u +C#

Now use #u=-3x# to finish:

#int e^(-3x) dx = -1/3 e^(-3x) +C#

For the definite integral, we don't need the #+C# (it subtracts out)

#int_0^3 e^(-3x) dx = -1/3 (e^(-3x) |_0^3) #

#= -1/3(e^(-3(3)) - e^(-3(0)))#

#= -1/3(e^(-9) - e^0)#.

#= -1/3(e^(-9) - 1)#

Now do whatever algebra you like.

I prefer:

#=1/3(1-1/e^9)# or #=1/3-1/(3e^9)# or #= (e^9-1)/(3e^9)#