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Integrating e^7x

  1. Mar 30, 2012 #1
    1. The problem statement, all variables and given/known data

    S e^7x


    2. Relevant equations

    no

    3. The attempt at a solution

    Ok so I am using U-substitution for this problem but I don't know what to do next.

    u = 7x, du = 7dx

    How do I integrate e^u*du?
     
  2. jcsd
  3. Mar 30, 2012 #2
    Try treating ∫eu du just as if it were ∫ex dx
     
  4. Mar 30, 2012 #3

    Mark44

    Staff: Mentor

    This is one of the easiest integrations!

    ##\int e^u~du = e^u + C##
     
  5. Mar 30, 2012 #4
    What happen's to du? Why does it dissapear in the solution?
     
  6. Mar 30, 2012 #5
    It disappears for the same reason when integrating something like ∫2x dx to get x2 + C
     
  7. Mar 30, 2012 #6
    I just want to point out, you are asking "how do I integrate"
    [tex]\int e^u \,du[/tex]

    Which implies a bit you MAY think that will give you the answer (you could have just omitted the 1/7 to be brief).

    But if you did miss it, since
    [tex]du =7dx \rightarrow dx = \frac{du}{7}[/tex]
    When you replace dx in the original integral with du/7 and 7x in the original integral with u, you get
    [tex]\int e^u \frac{du}{7}=\frac{1}{7} \int e^u \, du[/tex]
     
    Last edited: Mar 30, 2012
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