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Substitution Rule for Indefinite Integrals

  • Thread starter temaire
  • Start date
  • #1
279
0

Homework Statement


[tex]\int2e^-^7^xdx[/tex]


Homework Equations


None


The Attempt at a Solution



[tex](\frac{-2}{7})(\frac{e^-^7^x}{-7})+C[/tex]


This is as far as I can go, but the answer is:

[tex]\frac{-2e^-^7^x}{7}+C[/tex]
 

Answers and Replies

  • #2
1,752
1
How did you get two 1/7 terms?
 
  • #3
279
0
I took the antiderivative of [tex]e^-^7^x[/tex].
 
  • #4
1,752
1
I took the antiderivative of [tex]e^-^7^x[/tex].
So, why do you have 1/7 * 1/7? Just do it again from scratch and you'll probably see what you did wrong.
 
  • #5
1,752
1
[tex]2\int e^{-7x}dx[/tex]

[tex]-\frac 2 7\int -7e^{-7x}dx[/tex]

You don't need to divide by another -7. It's already in standard form!
 
  • #6
279
0
I've got it now, thanks.:smile:
 

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