The indefinite integral of 1/e^x is calculated as ∫ e^{-x} dx = -e^{-x} + C. The discussion highlights two methods for solving this integral: direct integration and substitution. The substitution method involves letting -x = t, which simplifies the integral to -∫ e^{t} dt = -e^{t} + C, and then substituting back to the original variable. The conversation emphasizes that while substitution is a valid approach, it may not be necessary for those familiar with basic integration techniques.
PREREQUISITESStudents studying calculus, educators teaching integration techniques, and anyone looking to strengthen their understanding of exponential integrals.
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.
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
\int e^{x}dx, for the other one\int e^{-x}dx, take the substitution -x=t, so now we have dx=-dt, now substitute back on the integral we get:
\int e^{t}(-dt)=-\int e^{t}dt=-e^t+c substituting back for the original variable we get -e^{-x}+c
JasonRox said:Why not just let u = e^-x?
No need for substitution here.