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amyna

  • one year ago

solve the intergal:

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  1. amyna
    • one year ago
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    |dw:1441503440519:dw|

  2. zepdrix
    • one year ago
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    Oh this should be similar to the e^(-x^2) one that you had earlier :)

  3. amyna
    • one year ago
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    i got the answer: -1/2 ln10 * 10^-x^2 not sure if thats correct because i don't really understand this problem!

  4. zepdrix
    • one year ago
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    Your ln(10) is is the denominator with the 2? But the 10^(-x^2) is not, correct?

  5. amyna
    • one year ago
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    yes

  6. zepdrix
    • one year ago
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    Hmm ya that looks correct! :) What is confusing you on this one? :O Integration of the exponential base 10?

  7. amyna
    • one year ago
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    ya thats what i don't get, like where the ln 10 came from

  8. zepdrix
    • one year ago
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    Do you remember how to differentiate something like this? :)\[\large\rm y=10^x\]\[\large\rm y'=?\]

  9. amyna
    • one year ago
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    no i don't

  10. zepdrix
    • one year ago
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    \[\large\rm y=10^x\]\[\large\rm \ln y=\ln 10^x\]\[\large\rm \ln y=x \ln 10\]Differentiating,\[\large\rm \frac{1}{y}y'=\ln 10\]\[\large\rm y'=y (\ln 10)\]\[\large\rm y'=10^x (\ln 10)\]So whenever we differentiate an exponential of base NOT e, we multiply by natural log of the base. When we integrate, we do the inverse of that, we divide by natural log of the base.\[\large\rm \int\limits 10^x dx=\frac{1}{\ln10}10^x\]

  11. amistre64
    • one year ago
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    works for base e as well

  12. amistre64
    • one year ago
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    y = e^x y' = e^x ln(e)

  13. zepdrix
    • one year ago
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    Yes, but unnecessary :) I guess that's why they don't show it for the derivative of e^x. I like seeing that extra step in the middle, helps to see the rule without forgetting about it.

  14. amistre64
    • one year ago
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    id rather learn one rule instead of special cases :)

  15. zepdrix
    • one year ago
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    You could also do this maybe Amy :O Recall that since the exponential base e, and natural log are inverse operations,\[\large\rm 10^x=e^{(\ln10^x)}\]Applying exponent rule gives,\[\large\rm =10^{x\color{orangered}{(\ln10)}}\]And maybe you remember how to integrate something of this form:\[\large\rm \int\limits e^{\color{orangered}{a}x}dx=\frac{1}{\color{orangered}{a}}e^{\color{orangered}{a}x}\]

  16. zepdrix
    • one year ago
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    Therefore,\[\large\rm \int\limits 10^{x}dx=\int\limits e^{\color{orangered}{(\ln10)}x}dx=\frac{1}{\color{orangered}{\ln10}}e^{\color{orangered}{(\ln10)}x}=\frac{1}{\ln10}10^x\]

  17. amyna
    • one year ago
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    Thank You so much! It make so much more sense now! :)

  18. zepdrix
    • one year ago
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    Ah I made a typo halfway through :( 10^(x(ln10)) should be e^(x(ln10))

  19. zepdrix
    • one year ago
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    Could you make sense of some of that? :O I know it was a lot to take in all at once lol

  20. amyna
    • one year ago
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    haha yes! Thank You!

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