B Derivative and integral of the exponential e^t

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Is the integral of e to the t the same as the derivative?
If I take the exponential function e^t and take the derivative, I think I get the same e^t. Even if I keep doing it over and over, second, third derivative, etc. My admittedly naive question, though, is this symmetric? Meaning...if I take the the integral of e^t, do I just get the reverse or do I have t deal with an infinity of constants because it is an indefinite integral?
 

PeroK

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Summary: Is the integral of e to the t the same as the derivative?

If I take the exponential function e^t and take the derivative, I think I get the same e^t. Even if I keep doing it over and over, second, third derivative, etc. My admittedly naive question, though, is this symmetric? Meaning...if I take the the integral of e^t, do I just get the reverse or do I have t deal with an infinity of constants because it is an indefinite integral?
Yes, an indefinite integral produces an equivalence class of functions, not a unique function.

This is important as you can see from the following example.

We want to find functions, ##f(t)## that satisfy ##f''(t) = e^t##. Clearly ##f(t) = e^t## is one such function. But, ##f(t) = e^t + At + B## is also a solution to this differential equation, where ##A, B## are any two constants.

Another way to look at this is to say that if you integrate ##f''(t) = e^t## you do not get a unique result.

Differentiation, on the other hand, always gives a unique function as a result.
 

HallsofIvy

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Yes, the derivative of [itex]e^x[/itex] is again [itex]e^x[/itex] and so the nth derivative, for all n, of [itex]e^x[/itex] is [itex]e^x[/itex] and all integrals of [itex]e^x[/itex] are again [itex]e^x[/itex]. In fact, that is why "e" is defined as it is. One can show that the derivative or [itex]a^x[/itex], for a any positive number, is [itex]C_aa^x[/itex] where "[itex]C_a[/itex]" is a constant (independent of x) that depends upon a. "e" is the unique number such that [itex]C_e= 1[/itex].
 

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