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ksaimouli

  • 2 years ago

integral

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  1. ksaimouli
    • 2 years ago
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    |dw:1360188674673:dw|

  2. ksaimouli
    • 2 years ago
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    @JamesJ

  3. ksaimouli
    • 2 years ago
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    |dw:1360189814146:dw|

  4. robtobey
    • 2 years ago
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    \[\text{The} \text{ derivative} \text{ of } 5^{x } \text { is}: \]\[5^x\text{ Log}[5] \]

  5. ksaimouli
    • 2 years ago
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    not sure i can do this because (5^x) (ln5) separate

  6. amoodarya
    • 2 years ago
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    use u=5^x

  7. ksaimouli
    • 2 years ago
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    then integral of u is (5^x)/ln5 right

  8. ksaimouli
    • 2 years ago
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    |dw:1360189966266:dw|

  9. ksaimouli
    • 2 years ago
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    if i use u substitution ln5 goes to top so no use

  10. Spacelimbus
    • 2 years ago
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    \[\Large \int 5^x\ln 5 dx \] ?? yes `

  11. ksaimouli
    • 2 years ago
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    yes

  12. Spacelimbus
    • 2 years ago
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    \[\Large \ln 5 \int 5^xdx = \ln 5 \int e^{\ln5x}dx \]

  13. Spacelimbus
    • 2 years ago
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    If I understand the integral correctly.

  14. Spacelimbus
    • 2 years ago
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    Just rewrite the number 5 as e^ln5 and then you have a regular exponential function (linear) you can substitute and evaluate.

  15. ksaimouli
    • 2 years ago
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    can u really do that manuplation

  16. Spacelimbus
    • 2 years ago
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    Of course, here's the proof: \[\Large a^b=c \\ \\ \Large b\log_aa= \log_ac\]that leads to : \[\Large b=\log_ac \] And back substitution gives: \[\Large a^{\log_ac}=c \] Just a different way of writing numbers, that's also the reason why a lot of people write the exponential decay with as an e function, rather then the regular y(t)=ba^t form

  17. Spacelimbus
    • 2 years ago
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    Integration result gives you \[\Large e^{\ln5 x}+C \] Deriving with respect to x: \[\Large \ln5 \cdot e^{\ln 5 x} = \ln 5 \cdot \left(e^{\ln5} \right)^x = \ln5 \cdot 5^x\]

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