2007-05-08 08:08:14 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It is (-1/3)e^(-3x) + C.
You can discover this using a u-substitution. Let u = -3x. Then du = -3 dx, or dx = du/-3. So we have
∫(e^(-3x)) dx = ∫e^u (du / -3) = (-1/3)∫(e^u) du.
Now, e^u is its own integral, so we get
(-1/3)∫(e^u) = (-1/3)e^u + C = (-1/3)e^(-3x) + C.
More generally, it is true that, for any nonzero real number k,
∫e^kx dx = (e^kx) / k + C.
2007-05-08 08:37:24 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The answer is:
(-1/3) (e^-3x) + C
You can check by taking the derivate of this, and you will get the original you started with.
2007-05-08 15:12:50 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Contant -3e^(-3x)
2007-05-08 15:12:14 · answer #3 · answered by gjmb1960 7 · 0⤊ 1⤋
-1/3e^-3x
2007-05-08 15:11:12 · answer #4 · answered by Anonymous · 0⤊ 0⤋
(e^(-3x+1)) / (-3x+1)
2007-05-08 15:12:05 · answer #5 · answered by Necrocell 1 · 0⤊ 1⤋