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invn177

  • 4 years ago

ok integral of INT/limits 1 to 0/ (e^x)^2

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  1. hbaldon
    • 4 years ago
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    rewrite?

  2. Ishaan94
    • 4 years ago
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    \[\int_0^1 e^{x^2}dx\]

  3. Ishaan94
    • 4 years ago
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    ah it is error function or something like that I can't do it

  4. Ishaan94
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=integration+e^%28x^2%29+from+0+to+1

  5. agreene
    • 4 years ago
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    yeah, i think its like 1/2 sqrt(pi)*error(imaginary) or some such

  6. hbaldon
    • 4 years ago
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    yep. that's what my calcu says.. 1.462651746 xD

  7. mathTalk
    • 4 years ago
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    what is this now ?

  8. invn177
    • 4 years ago
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    nonono, (e^x)(e^x) or (e^x)^2, integrate that from 1 to 0

  9. invn177
    • 4 years ago
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    And I have a computer right in front of me as well as a ti-84+ so I dont want your calculator or wolfram alpha answers please :)

  10. agreene
    • 4 years ago
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    Oh, that is much simpler. (e^x)^2 = e^2x so: \[\int\limits_{0}^{1}e^{2x}dx=\frac{1}{2}e^2x\] take it to the limits: [1/2*e^(2)]-[1/2*e^(0)]=1/2e^2-1/2 factor it and you have: \[\frac{1}{2}(e^2-1)\]

  11. invn177
    • 4 years ago
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    oh you sure that (e^x)^2=e^2x?

  12. agreene
    • 4 years ago
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    yes, that is one of the properties of exponents

  13. invn177
    • 4 years ago
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    fsho

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