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EsKiMo |
Someone's signature was a challenge about an integral... Integral of e^x , I think The solution is simple, (e^x) / Log(e) + C or simply e^x + C I don't remember who was he, but the answer is here Bye |
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Edited by EsKiMo on 07.12.2003 12:50:59 | |||
07.12.2003 00:16:03 |
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Jagori |
Isn't the integral of e^x just e^x ? (plus some arbitrary constant) I think the challenge in the signature was to take the integral of e^x², but I may be mistaken as I haven't seen that sig in a while |
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Edited by Jagori on 07.12.2003 02:22:01 | |||
07.12.2003 02:20:02 |
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SA007 |
jeah, because e is chosen in that way that log(e) = 1, so you can leave away log(e), but you dont have to. |
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07.12.2003 08:32:19 |
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DigitalAcid |
I think it was lefeu... |
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07.12.2003 16:15:37 |
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