jmankickin jmankickin
  • 13-11-2016
  • Mathematics
contestada

6.4e^(−3x + 6) dx
take the integral

Respuesta :

bartdrinksmalk
bartdrinksmalk bartdrinksmalk
  • 13-11-2016
First, let's write 6.4 as a fraction to make it a bit easier to work with.
[tex]\int\ {\frac{32}{5}e^{-3x+6}} \, dx[/tex]
[tex]=\frac{32}{5} \int\ {e^{-3x+6}} \, dx[/tex]
Use u-substitution:
[tex]u=-3x+6[/tex]
[tex]du=-3dx[/tex]
So we now have
[tex]\frac{32}{5}*-\frac{1}{3} \int {-3e^{-3x+6}} \, dx[/tex]
[tex]=-\frac{32}{15} \int {e^{u}} \, du[/tex]
[tex]=-\frac{32}{15}e^{u}+C[/tex]
[tex]=-\frac{32}{15}e^{-3x+6}+C[/tex]

(If you want the answer to have a decimal, then [tex]-\frac{32}{15}[/tex] is -2.133333...)
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