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- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,678

- Mar 1, 2012

- 655

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

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 ?

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- #3

- Jan 31, 2012

- 2,678

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

- Mar 1, 2012

- 655

ok ... continue with part (c)

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- #5

- Jan 31, 2012

- 2,678

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.....