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 April 11th, 2017, 08:20 PM #1 Member   Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0 Find the area of the region between the curve I've asked a similar problem on here already (still waiting for a reply) but it has something new to this one, the horizontal part of it isnt it just 0?
 April 11th, 2017, 08:38 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond Hi Bobbyjoe and welcome to the forum. You need to evaluate $$8\int_2^4\ln(x)\,dx$$ Can you proceed?
April 11th, 2017, 08:42 PM   #3
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Quote:
 Originally Posted by greg1313 Hi Bobbyjoe and welcome to the forum. You need to evaluate $$8\int_2^4\ln(x)\,dx$$ Can you proceed?
so am I not suppose to acknowledged the horizontal axis?

 April 11th, 2017, 08:51 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond The definite integral I gave will give the area between $8\ln(x)$ and the $x$-axis. The length of the curve of $8\ln(x)$ is above $y=0$ for the interval $[2,4]$.
April 11th, 2017, 08:53 PM   #5
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Quote:
 Originally Posted by greg1313 The definite integral I gave will give the area between $8\ln(x)$ and the $x$-axis. The length of the curve of $8\ln(x)$ is above $y=0$ for the interval $[2,4]$.
Oh okay, I think I understand! Thank you so much!

April 12th, 2017, 05:53 AM   #6
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Quote:
 Originally Posted by Bobbyjoe so am I not suppose to acknowledged the horizontal axis?
The horizontal axis is y= 0. You are integrating 8 ln(x)- 0= 8ln(x).

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