Area Calculation for y = ||e^x - 1| - 1|

[phospho]

Q. Show that the area between the positive x-axis, the y-axis and the curve $y = ||e^x - 1| - 1|$ is ln4 - 1

I've drawn the curve:

http://gyazo.com/cfd52af0f82e0e7d6b063681a73de45a

I notice for x < 0 (as I drew e^x to start out with, that's how I noticed it):
y = e^xfor x>ln(2) y = e^x - 2 (again, as I drew it before hand)

I can't seem to see what y will be for for 0≤x≤ln(2), as all my other notices are because I drew them beforehand.

Could anyone explain how I can split up y accordingly for x <0 for 0≤x≤ln(2) x > ln(2) ?

 

[mfb]

For 0≤x≤ln(2), which range does e^x cover? In particular, what is its smallest value? Based on that, can you get rid of the inner modulus? In the same way, you can get replace the outer modulus.

 

[SammyS]

Q. Show that the area between the positive x-axis, the y-axis and the curve $y = ||e^x - 1| - 1|$ is ln4 - 1

I've drawn the curve:

http://gyazo.com/cfd52af0f82e0e7d6b063681a73de45a

I notice for x < 0 (as I drew e^x to start out with, that's how I noticed it):
y = e^x


for x>ln(2) y = e^x - 2 (again, as I drew it before hand)

I can't seem to see what y will be for for 0≤x≤ln(2), as all my other notices are because I drew them beforehand.

Could anyone explain how I can split up y accordingly for x <0 for 0≤x≤ln(2) x > ln(2) ?

Use the definition of absolute value, twice.
$$|e^x-1| =<br /> \begin{cases}<br /> e^x-1 & \text{if } e^x \geq 1 \quad \text{i.e. } x\ge 0 \\ \\<br /> 1-e^x & \text{if } e^x < 1 \quad \text{i.e. } x < 0<br /> \end{cases}$$
$$||e^x-1|-1| =<br /> \begin{cases}<br /> |e^x-1|-1 & \text{if } |e^x-1| \geq 1 \quad \text{i.e. } x\ge \ln(2) \\ \\<br /> 1-|e^x-1| & \text{if } |e^x-1| < 1 \quad \text{i.e. } x < \ln(2)<br /> \end{cases}$$
So you want x < ln(2) which gives $\displaystyle ||e^x-1|-1| =1-|e^x-1| \,,\$ but also x > 0 which tells you that $\displaystyle |e^x-1|=e^x-1 \ .$

Put those together.

 

[phospho]

thank you!