megannicole51
  • megannicole51
Find the area between y=ln(x) and y=ln(x^2) for 1 is greater than or equal to x and greater than or equal to 2?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1378343183294:dw|
megannicole51
  • megannicole51
so the integral from 1 to 2 of ln(x)dx is the answer? and thank you for drawing and writing that out it makes sense! i completely forgot all of this is calc 1 haha!
anonymous
  • anonymous
So your dealing with: \[A=\int^2_1{ln(x^2)}dx-\int^2_1{ln(x)}dx\] Lets break it up like so: \[A=\gamma-\psi\] And solve each. So we have: \[\gamma=\int^2_1{ln(x^2)}dx=2\int^2_1{ln(x)}\] \[=2\left.(xln(x)-x)\right|^2_1=2[(2ln(x)-2)-(ln(x)-1)]=2[ln(x)-1]=2ln(x)-2\]

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megannicole51
  • megannicole51
i cant see the end of ur solution...it got cut off
anonymous
  • anonymous
Haha im getting there
anonymous
  • anonymous
how does \[\int\limits \ln x\, dx = x?\]
megannicole51
  • megannicole51
it doesn't?
anonymous
  • anonymous
Noo...\[\int{ln(x)}=xln(x)-x+C\]
anonymous
  • anonymous
lol...
megannicole51
  • megannicole51
wait were u asking me or KeithAfasCalcLover
anonymous
  • anonymous
KeithAfasCalcLover
megannicole51
  • megannicole51
lol my bad.....wait so wat ur first post was made sense but is that it? pgpilot326
anonymous
  • anonymous
No theres more.
anonymous
  • anonymous
\[\ln x^{2} = \ln x +\ln x=2\ln x\]
anonymous
  • anonymous
so all you get is \[\int_{1}^{2} \ln x\, dx\]
megannicole51
  • megannicole51
so i dont have to solve for that integral? like plug in 1 and 2 for x? i believe that was like the fundamental theorem of calc right?
anonymous
  • anonymous
you have to integrate ln x by parts. do you know how to do that?
megannicole51
  • megannicole51
somewhat but not very well...
megannicole51
  • megannicole51
my professor doesnt speak english well so its been difficult to learn this stuff
anonymous
  • anonymous
Whoops yeah I was going the long way, @pgpilot326 is correct so in the end, you have \[\int^2_1{ln(x) dx}=[xln(x)-x]|^2_1=[2ln(2)-2]-[ln(1)-1]=2ln(2)-1\]
anonymous
  • anonymous
And that can be difficult haha
anonymous
  • anonymous
Acording to the Fundamental Theorem of calculus, : \[\int^b_a{f(x)dx}=F(b)-F(a)\]
anonymous
  • anonymous
Where \(F'(x)=f(x)\)
anonymous
  • anonymous
|dw:1378344197954:dw|
anonymous
  • anonymous
sorry... difficult to write with a mouse! squirmy little bstard
megannicole51
  • megannicole51
lol its okay
anonymous
  • anonymous
Haha learning how to type with LaTex is easier so like codes to render nice equations on the screen
anonymous
  • anonymous
just takes longer right now
anonymous
  • anonymous
So in general, \[\int{ln(x)dx}=xln(x)-x+C\]
anonymous
  • anonymous
are you good @megannicole51 ?
megannicole51
  • megannicole51
yup it makes sense thank you!
anonymous
  • anonymous
you're welcome!
anonymous
  • anonymous
So in our case, \[\int^2_1{ln(x)dx}=[xln(x)-x]|_{x=2}-[xln(x)-x]|_{x=1}\]

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