Find the Area Of the Region

KEYWEST17

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I plugged the equations into the graphing calculator and can see the areas to evaluate but I am stuck.
 

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I assume the shaded region is the area in question?.

Find the area under \(\displaystyle e^{2x}\) from 0 to 1:

\(\displaystyle \int_{0}^{1}e^{2x}dx\)

Find the area under \(\displaystyle e^{6x}\) from \(\displaystyle -\infty \;\ \text{to} \;\ 0\):

\(\displaystyle \int_{-\infty}^{0}e^{6x}dx\)

Add them up.

This is assuming my assessment is correct and I have the correct region as in the diagram.

But, if you need the region between the two e functions: \(\displaystyle \int_{-\infty}^{0}\left[e^{2x}-e^{6x}\right]dx\)
 

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KEYWEST17 said:
I plugged the equations into the graphing calculator and can see the areas to evaluate but I am stuck.

KEYWEST17,

you are also to include y = 0, as another curve to form the region,
if it is supposed to be the figure drawn by galactus.
 
No, I am sorry to say, it does not.

Where did the negative come from?.

The region is above the x-axis.
 
I am unsure of what region you are looking for.

Is it \(\displaystyle \int_{0}^{1}e^{2x}dx\)?.

Or

\(\displaystyle \int_{-\infty}^{0}\left(e^{2x}-e^{6x}\right)dx\)?.

Or both?.

Or something else?.

The problem may indicate both.

\(\displaystyle \int_{-\infty}^{0}\left(e^{2x}-e^{6x}\right)dx+\int_{0}^{1}e^{2x}dx\)

Show us what you're doing so we can pinpoint the trouble.
 
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