e^(-x²) has no elementary antiderivative -- which is not to say that it has no antiderivative. It does, and this antiderivative is [0, x]∫e^(-t²) dt. However this cannot be written using any finite combination of powers, roots, exponents, logs, or trigonometric functions. If you need to compute this for some reason, you can find a taylor series for it as follows: note that e^x = [k=0, ∞]∑x^k/k!, so e^(-x²) = [k=0, ∞]∑(-1)^k x^(2k)/k!. Integrating this term by term yields that [0, x]∫e^(-t²) dt = [k=0, ∞]∑(-1)^k x^(2k+1)/((2k+1)k!)
2007-05-02 12:37:25
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answer #1
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answered by Pascal 7
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That looks familiar, kinda looks like the bell curve equation. I tried integrating by parts, but couldn't figure it out. Then I plugged it into my Ti-89 and it gave me the same function out. Let me know, if you find an answer.
2007-05-02 12:28:38
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answer #2
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answered by Anonymous
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no way, there should be a x somewhere!!
check the whole problem once more
2007-05-02 12:28:19
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answer #3
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answered by nelaq 4
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