∫(-1 to 0) e^-x dx = ....?

Question

its ethier 1-e or (1-e)/e... not sure wat one is right tho cause we never actually learned integrals in calc yet... but those are two answers i came up with like to see if anyone can tell me if one is right

Answer 1

e - 1

Integrating gives -e^-x.
Evaluating gives (-e^0) - (-e^(+1)) = -1 + e = e - 1

Answer 2

The derivative of e^-x = - e^-x

The integral from -e^-x is e^-x

So, - int e^-x = e^-x

int- e^-x = - int e^-x

And int e^-x = - e^-x

Now you consider the integrations limits

e^-x, evaluated in 0 = 1

e^-x, evaluated in -1 = e^1

So, the result is - e^-0 - (-e^-1) = -1 - (-e) = e-1

Ana

Answer 3

I = ∫ e^(- x).dx between limits - 1 to 0
I = - [ e^(- x) ] between limits - 1 to 0
I = - [ 1 - e^1 ]
I = e - 1