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virtus Group Title

integrate y=2^x from x=0 to x=3

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    \[\frac d{dx}a^x=a^x\ln a\]so then what is the integral?

    • 2 years ago
  2. virtus Group Title
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    a^x +c ?

    • 2 years ago
  3. virtus Group Title
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    actually i don't know hahaha

    • 2 years ago
  4. Mikey Group Title
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    The indefinite integral of a^x is a^x/lna + C

    • 2 years ago
  5. TuringTest Group Title
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    \[\int a^x\ln adx=a^x\]so\[\int a^xdx=\frac{a^x}{\ln a}\]+C if it is indefinite, but yours is definite anyway...

    • 2 years ago
  6. virtus Group Title
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    turning test can you please show me how this was derived. Thank you very much!

    • 2 years ago
  7. TuringTest Group Title
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    do you need me to prove that\[\frac d{dx}a^x=a^x\ln a\]? that would be the most thorough way to start

    • 2 years ago
  8. virtus Group Title
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    yes please!

    • 2 years ago
  9. TuringTest Group Title
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    we can prove it with logarithmic differentiation\[y=a^x\]taking the natural of of both sides\[\ln y=\ln(a^x)\]\[\ln y=x\ln a\]now differentiate implicitly\[\frac{y'}y=\ln a\]\[y'=y\ln a=a^x\ln a\]so what does this tell us about the antiderivative?

    • 2 years ago
  10. TuringTest Group Title
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    \[y'=a^x\ln a\iff y=a^x\]so\[\int a^x\ln a=a^x\]now we can use this info and do your integral with a u-substitution

    • 2 years ago
  11. TuringTest Group Title
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    watch the sub carefully:\[\int a^xdx\]\[u=a^x\]\[du=a^x\ln a dx\to\frac{du}{\ln a}=a^xdx\]subbing in our expresison for a^xdx we get\[\frac1{\ln a}\int du=\frac u{\ln a}+C=\frac{a^x}{\ln a}+C\]

    • 2 years ago
  12. TuringTest Group Title
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    any questions about that?

    • 2 years ago
  13. virtus Group Title
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    THANK YOU A MILLION TIMES OVER turning test! greatly appreciated

    • 2 years ago
  14. TuringTest Group Title
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    anytime :D

    • 2 years ago
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