How to Find the Point of Intersection Between ln x and 5-x?

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SUMMARY

The discussion focuses on finding the point of intersection between the functions $\ln x$ and $5-x$, which is determined to be at $x=3.69344$. The area between the curves is calculated using the integral $\displaystyle R = \int_1^a \ln{x} \, dx + \int_a^5 5-x \, dx$, where $a$ represents the intersection point. Additionally, the volume of the solid formed by revolving the area around an axis is expressed as $\displaystyle V = \int_1^a (\ln{x})^2\, dx + \int_a^5 (5-x)^2 \, dx$. The discussion also explores alternative formulations of the integral to find the area and volume.

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karush
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Ok this might take a while...
but first find point of intersection $\ln x=5-x$
which calculates to $x=3.69344$ which maybe there is more simpler approach
 
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this is a calculator active problem ...

$\displaystyle R = \int_1^a \ln{x} \, dx + \int_a^5 5-x \, dx$
where $a$ is the x-value of the intersection.

or ...
$\displaystyle R = \int_0^b (5-y) - e^y \, dy$
where $b$ is the y-value of the intersection.

can you set up the volume by similar cross-section integral ?
 
$\displaystyle V = \int_1^a (\ln{x})^2\, dx + \int_a^5 (5-x)^2 \, dx$
 
ok ... continue with part (c)
 
skeeter said:
ok ... continue with part (c)
if we chanhge b to k
$\displaystyle \int_0^k (5-y) - e^y \, dy = \dfrac{1}{2} A$
then solve for k y was derived previous
anyway...
 

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