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amyna
 one year ago
solve the intergal:
amyna
 one year ago
solve the intergal:

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Oh this should be similar to the e^(x^2) one that you had earlier :)

amyna
 one year ago
Best ResponseYou've already chosen the best response.0i got the answer: 1/2 ln10 * 10^x^2 not sure if thats correct because i don't really understand this problem!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Your ln(10) is is the denominator with the 2? But the 10^(x^2) is not, correct?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Hmm ya that looks correct! :) What is confusing you on this one? :O Integration of the exponential base 10?

amyna
 one year ago
Best ResponseYou've already chosen the best response.0ya thats what i don't get, like where the ln 10 came from

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Do you remember how to differentiate something like this? :)\[\large\rm y=10^x\]\[\large\rm y'=?\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\large\rm y=10^x\]\[\large\rm \ln y=\ln 10^x\]\[\large\rm \ln y=x \ln 10\]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
 one year ago
Best ResponseYou've already chosen the best response.0works for base e as well

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0y = e^x y' = e^x ln(e)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0id rather learn one rule instead of special cases :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4You 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
 one year ago
Best ResponseYou've already chosen the best response.4Therefore,\[\large\rm \int\limits 10^{x}dx=\int\limits e^{\color{orangered}{(\ln10)}x}dx=\frac{1}{\color{orangered}{\ln10}}e^{\color{orangered}{(\ln10)}x}=\frac{1}{\ln10}10^x\]

amyna
 one year ago
Best ResponseYou've already chosen the best response.0Thank You so much! It make so much more sense now! :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Ah I made a typo halfway through :( 10^(x(ln10)) should be e^(x(ln10))

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Could you make sense of some of that? :O I know it was a lot to take in all at once lol
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