amyna
  • amyna
solve the intergal:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amyna
  • amyna
|dw:1441503440519:dw|
zepdrix
  • zepdrix
Oh this should be similar to the e^(-x^2) one that you had earlier :)
amyna
  • amyna
i got the answer: -1/2 ln10 * 10^-x^2 not sure if thats correct because i don't really understand this problem!

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zepdrix
  • zepdrix
Your ln(10) is is the denominator with the 2? But the 10^(-x^2) is not, correct?
amyna
  • amyna
yes
zepdrix
  • zepdrix
Hmm ya that looks correct! :) What is confusing you on this one? :O Integration of the exponential base 10?
amyna
  • amyna
ya thats what i don't get, like where the ln 10 came from
zepdrix
  • zepdrix
Do you remember how to differentiate something like this? :)\[\large\rm y=10^x\]\[\large\rm y'=?\]
amyna
  • amyna
no i don't
zepdrix
  • zepdrix
\[\large\rm y=10^x\]\[\large\rm \ln y=\ln 10^x\]\[\large\rm \ln y=x \ln 10\]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.\[\large\rm \int\limits 10^x dx=\frac{1}{\ln10}10^x\]
amistre64
  • amistre64
works for base e as well
amistre64
  • amistre64
y = e^x y' = e^x ln(e)
zepdrix
  • zepdrix
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.
amistre64
  • amistre64
id rather learn one rule instead of special cases :)
zepdrix
  • zepdrix
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:\[\large\rm \int\limits e^{\color{orangered}{a}x}dx=\frac{1}{\color{orangered}{a}}e^{\color{orangered}{a}x}\]
zepdrix
  • zepdrix
Therefore,\[\large\rm \int\limits 10^{x}dx=\int\limits e^{\color{orangered}{(\ln10)}x}dx=\frac{1}{\color{orangered}{\ln10}}e^{\color{orangered}{(\ln10)}x}=\frac{1}{\ln10}10^x\]
amyna
  • amyna
Thank You so much! It make so much more sense now! :)
zepdrix
  • zepdrix
Ah I made a typo halfway through :( 10^(x(ln10)) should be e^(x(ln10))
zepdrix
  • zepdrix
Could you make sense of some of that? :O I know it was a lot to take in all at once lol
amyna
  • amyna
haha yes! Thank You!

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