What is the Indefinite Integral of 1/e^x?

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In summary, the indefinite integral of 1/e^x is -e^-x + c. This can be found by substituting -x for t and then substituting back for the original variable, or by letting u = e^-x. This approach may make it easier to understand for those who are not familiar with integrating e^-x.
  • #1
mathguyz
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Does anyone here know what the indefinite integal of 1/e^x is?

Im an new member and, this is my first post.
 
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  • #2
mathguyz said:
Does anyone here know what the indefinite integal of 1/e^x is?

Im an new member and, this is my first post.

i guess u have posted on the wrong forum, but anyways!

[tex]\int e^{-x} dx= -e^{-x}+c[/tex]
 
  • #3
here it is how you might want to think about it: i am assuming you know how to integrate this, since it is only a tabelar integral
[tex]\int e^{x}dx[/tex], for the other one[tex]\int e^{-x}dx[/tex], take the substitution -x=t, so now we have dx=-dt, now substitute back on the integral we get:
[tex]\int e^{t}(-dt)=-\int e^{t}dt=-e^t+c [/tex] substituting back for the original variable we get [tex] -e^{-x}+c[/tex]
 
  • #4
sutupidmath said:
here it is how you might want to think about it: i am assuming you know how to integrate this, since it is only a tabelar integral
[tex]\int e^{x}dx[/tex], for the other one[tex]\int e^{-x}dx[/tex], take the substitution -x=t, so now we have dx=-dt, now substitute back on the integral we get:
[tex]\int e^{t}(-dt)=-\int e^{t}dt=-e^t+c [/tex] substituting back for the original variable we get [tex] -e^{-x}+c[/tex]

Why not just let u = e^-x?

No need for substitution here.
 
  • #5
JasonRox said:
Why not just let u = e^-x?

No need for substitution here.

yeah, i know, but since the op was not able to integrate e^-x, i was assuming that approaching this problem like i did, it would make it easier for him/her to understand!
 

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