anonymous
  • anonymous
|dw:1333526535476:dw|
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Diyadiya
  • Diyadiya
Drawings wont work in the question :)
anonymous
  • anonymous
eish k let me write it then
Diyadiya
  • Diyadiya
ok

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anonymous
  • anonymous
Determine integral (with top part of it with e^3 and the bottom part of it e^2) of: In(In w)/wlog3w (this 3 its small and just slight below log) is
Diyadiya
  • Diyadiya
You can draw or use the Equation editor While replying :) Hope someone will help you out!
anonymous
  • anonymous
\[\int\limits_{e^2}^{e^3}In(Inw)/wlog3w dw\]
anonymous
  • anonymous
i tried to type it
anonymous
  • anonymous
but the 3 in log is slight below the log, its not 3w, the 3 its slightly belowthe log
dumbcow
  • dumbcow
oh ok so log base 3
anonymous
  • anonymous
yes
dumbcow
  • dumbcow
ok first get rid of that log by using change of base \[\rightarrow \log_{3}w = \frac{\ln w}{\ln 3} \] simplify the integral \[\rightarrow \ln 3\int\limits_{?}^{?}\frac{\ln (\ln w)}{w \ln w}dw\]
dumbcow
  • dumbcow
then make substitution u = ln w du = 1/w dw --> dw = w du \[\rightarrow \ln 3\int\limits_{?}^{?} \frac{\ln u}{u}du\]
dumbcow
  • dumbcow
then use integration by parts: x = ln u dv = 1/u dx = 1/u v = ln u \[\int\limits_{?}^{?} \frac{\ln u}{u}du = \ln^{2}u - \int\limits_{?}^{?} \frac{\ln u}{u}du\] \[2\int\limits_{?}^{?} \frac{\ln u}{u}du = \ln^{2} u\] \[\int\limits_{?}^{?} \frac{\ln u}{u}du = \frac{\ln^{2} u}{2}\] sub back in for u and put limits in \[\rightarrow \ln 3(\frac{\ln^{2} (\ln w)}{2}) from e^{3} \to e^{2}\]
dumbcow
  • dumbcow
i get about 0.4
anonymous
  • anonymous
thank you i see, i never knew about intergration by parts
anonymous
  • anonymous
do you mind explaining to me how it works
dumbcow
  • dumbcow
welcome, sure anytime you have a product of 2 functions, you can use integration by parts \[\int\limits_{}^{}f(x)g(x) dx\] assign either f or g as u and the other as dv, where u and v are functions of x this leads to the equation: \[\int\limits_{}^{}u dv = uv -\int\limits_{}^{}v du\] so to use the equation you must find du and v lets say u = f(x) then du = f'(x) then dv = g(x) and v = integral g(x)
dumbcow
  • dumbcow
here is an example: \[\int\limits_{}^{}xe^{x}dx\] u = x dv = e^x du = dx v = e^x \[\int\limits\limits_{}^{}xe^{x}dx = xe^{x}-\int\limits_{}^{}e^{x} dx\]
anonymous
  • anonymous
k even if both function are are in the division format? e.g. \[e^xdivx\]
anonymous
  • anonymous
u can still use the the formula?
dumbcow
  • dumbcow
yes, but one of the functions would be 1/x u = e^x dv = 1/x du = e^x v = ln x notice this will not help much though actually integral e^x/x can't be done using regular methods but in general yes even division can be represented as product of 2 functions
anonymous
  • anonymous
oh k see i get it, thank you, Be blessed

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