i should know this, but i forgot what it is:
what is the antiderivative of e^-x dx?
2007-12-02 04:55:21 · 6 answers · asked by cornz000 1 in Science & Mathematics ➔ Mathematics
Oh, thats easy. Its -e^-x+c. If you have doubt, follow these steps
Take u=-x
Then, du/dx=-1
dx=-du
Substituting,
{e^-xdx={e^u.-du
=-e^u+c
=-e^-x+c
Ok, but don't forget to give me best answer if you get it.
2007-12-02 05:03:37 · answer #1 · answered by Faheem 4 · 0⤊ 0⤋
try to take derivative of e^(-x).
That is
(e^(-x)) ' = - e^(-x)
Oops, you have the negative. How do you solve this stampede?
just take the negative of e^(-x) and you get
(-e^(-x))' = e^(-x)
Hence integral of e^(-x) is -e^(-x)
2007-12-02 13:01:47 · answer #2 · answered by Theta40 7 · 0⤊ 0⤋
Let u = - x
du = - dx
I = - â« e^u du
I = - e^u + C
I = - e^(-x) + C
2007-12-04 18:09:21 · answer #3 · answered by Como 7 · 1⤊ 0⤋
-ea-x
2007-12-02 13:08:12 · answer #4 · answered by Anonymous · 0⤊ 0⤋
-e^-x
2007-12-02 12:58:31 · answer #5 · answered by MYRA M 3 · 0⤊ 0⤋
-e^-x
2007-12-02 12:57:55 · answer #6 · answered by KVC 3 · 1⤊ 0⤋