2007-04-30 16:21:35 · 7 answers · asked by itsSCIENCE 2 in Science & Mathematics ➔ Mathematics
I = ∫ e^(2x).dx
Let u = 2x
du/dx = 2
dx = du / 2
I = (1/2).∫ e^(u).du
I = (1/2).e^(u) + C
I = (1/2).e^(2x) + C
2007-04-30 22:08:08 · answer #1 · answered by Como 7 · 0⤊ 0⤋
I believe its 1/2 e^2x. Taking that derivative gives 1/2 e^2x(2) = e^2x.
2007-04-30 23:24:58 · answer #2 · answered by spmdrumbass 4 · 0⤊ 0⤋
integral e^(2x) dx
Let u = 2x. Then du/dx = 2 ==> dx = du/2
integral e^u du/2 = (1/2) integral e^u du = (1/2)e^u + C = (1/2)e^(2x) + C
2007-04-30 23:25:12 · answer #3 · answered by DavidK93 7 · 2⤊ 0⤋
e^u du = e^u
thus integral [e^2x]dx = 1/2[e^2x]2dx = 1/2e^2x
2007-04-30 23:26:20 · answer #4 · answered by mikedotcom 5 · 0⤊ 0⤋
(1/2)e^(2x) + c
2007-04-30 23:25:19 · answer #5 · answered by gudspeling 7 · 2⤊ 0⤋
(e^2x)/2 + c
2007-04-30 23:26:03 · answer #6 · answered by Superconductor 3 · 0⤊ 0⤋
(e^2x)/2. Verify this for yourself that e^(nx) has antiderivative [e^(nx)]/n for all n, using u-substitution.
2007-04-30 23:25:50 · answer #7 · answered by Anonymous · 0⤊ 0⤋