- #1

karush

Gold Member

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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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- Thread starter karush
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- #1

karush

Gold Member

MHB

- 3,267

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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

- #2

skeeter

- 1,104

- 1

$\displaystyle R = \int_1^a \ln{x} \, dx + \int_a^5 5-x \, dx$

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 ?

- #3

karush

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$\displaystyle V = \int_1^a (\ln{x})^2\, dx + \int_a^5 (5-x)^2 \, dx$

- #4

skeeter

- 1,104

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ok ... continue with part (c)

- #5

karush

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if we chanhge b to kok ... continue with part (c)

$\displaystyle \int_0^k (5-y) - e^y \, dy = \dfrac{1}{2} A$

then solve for k y was derived previous

anyway...

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