2006-12-22 01:18:17 · 12 answers · asked by fm02 2 in Science & Mathematics ➔ Mathematics
-e^(-x)+c
where c is the constant of integration.
2006-12-22 01:20:41 · answer #1 · answered by Som™ 6 · 1⤊ 0⤋
Try substitution
Let u = -x
du=-dx so -du=dx
integral e^u * -du =
- integral e^u du = - e^u +C = -e^(-x) +C
Comment: I do not memorize special cases. I use substitution because it's an application of the Chain Rule for integration. You have the composition of two function e^x and -x.
2006-12-22 09:21:39 · answer #2 · answered by Professor Maddie 4 · 1⤊ 1⤋
Why do you need to do substitution (@professor..)?
integral e^(-x) dx = -e^(-x)+c
where c is the integration constant
2006-12-22 09:32:42 · answer #3 · answered by l_kur 5 · 0⤊ 0⤋
Integral of a*e^(bx) dx is (a/b)e^(bx) + C
Therefore integral of e^(-x) dx is -e^(-x) + C
2006-12-22 09:31:12 · answer #4 · answered by Kevin 2 · 0⤊ 0⤋
=e^-x/-1
=-1/e^x
2006-12-22 11:41:31 · answer #5 · answered by openpsychy 6 · 0⤊ 1⤋
Please visit web site www.mathworld.wolfram.com and you will get what you want and remember me after you find your this or any other solution to the problem, enjoy and solve your problems with the help this site gives
2006-12-22 09:32:50 · answer #6 · answered by Harinder S. Johal 7 · 0⤊ 0⤋
e^(-x)/(-1)
2006-12-22 09:30:15 · answer #7 · answered by mac 1 · 0⤊ 0⤋
â«e^-x dx=-e^-x +c
2006-12-22 11:38:30 · answer #8 · answered by mu_do_in 3 · 0⤊ 0⤋
-e^(-x)
2006-12-24 01:53:18 · answer #9 · answered by bubbly 2 · 0⤊ 0⤋
-e^(-x)+c
2006-12-23 02:57:34 · answer #10 · answered by sidd the devil 2 · 0⤊ 0⤋