Question

quick question. can someone explain to me e^-x when integrated is -e^-x. thanks!

Answer 1

Because of two rules. The first is the chain rule, that says that:
$$
\bigg(f \big[ g(x) \big] \bigg)' = f'\big[g(x)\big] \cdot g'(x)
$$In our case:
$$
f(x)=e^x \\
g(x)=-x\\
f\big[g(x)\big]=e^{-x} \\
(e^{-x})'=e^{-x}\cdot(-x)'=-e^{-x}
$$
The other rule is basically the factor rule, which could be though as a special case of the product* (fixed, wrote 'chain' at first by accident =S) rule, where one of the functions is constant. In short:
$$(a \cdot f(x))'=a \cdot f′(x)$$So:
$$(-e^{-x})'=-(e^{-x})'=-(−e^{-x})=e^{-x}$$
And therefore the anti-derivative of \(e^{-x}\) is \(-e^{-x}\)
If you understand why the rules are correct then it should make sense =)
Hope it's clear

Answer 2

holy pellet thank you for that explanation!

Answer 3

Sure no problem =)