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virtus

  • 3 years ago

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

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  1. TuringTest
    • 3 years ago
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    \[\frac d{dx}a^x=a^x\ln a\]so then what is the integral?

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

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

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

  5. TuringTest
    • 3 years ago
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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...

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

  7. TuringTest
    • 3 years ago
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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

  8. virtus
    • 3 years ago
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    yes please!

  9. TuringTest
    • 3 years ago
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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?

  10. TuringTest
    • 3 years ago
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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

  11. TuringTest
    • 3 years ago
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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\[\frac1{\ln a}\int du=\frac u{\ln a}+C=\frac{a^x}{\ln a}+C\]

  12. TuringTest
    • 3 years ago
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    any questions about that?

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

  14. TuringTest
    • 3 years ago
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    anytime :D

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