Apc.4.1.13 find Region

[karush]

i

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

 

[skeeter]

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 ?

 

[karush]

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

 

[skeeter]

ok ... continue with part (c)

 

[karush]

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...