Find the Area of the region

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KEYWEST17

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Decide whether to integrate with respect to x or y. Find the area of the region: y=e^2^x , y=e^6^x , x=1

So far I made (1/6e^6 - (1/6)) - (1/2 e^2 - (1/2))

(1/6e^6 - 1/6) - (1/2 e^2 + 1/2)

I simplified and got 1/6e^6 - 1/2e^2 -2/3

For some reason this answer is coming up wrong. Did I miss a simple step?
 
You don't seem to have cared in which direction you found the integral. Is it dy or dx. Decide, state it clearly, and give it another go.

Are you sure your three specifications actually define a region?

Based on what appears to be your antiderivatives for purposes of integration, I no longer can tell what the original function bounds were. "e" is NOT a variable.

Did you really mean this? \(\displaystyle y\;=\;e^{2^{x}}\)
 
Okay.

Now tell me the antiderivative of \(\displaystyle e^{x}\)
 
Fair enough. Then where did this come from... 1/6e^6

Maybe I'm not understanding your notation. How did you find the antiderivative of \(\displaystyle e^{2^{x}}\)?

Anyway, the first question would be the direction. Are you sure dx looks better?
 
tkhunny said:
Did you really mean this? \(\displaystyle y\;=\;e^{2^{x}}\)
Click to expand...

KEYWEST17 said:
Thats how I meant to write it.
Click to expand...

No, KEYWEST17, you did not mean it that way. They have to be \(\displaystyle y \ = \ e^{2x} \ \ and \ \ y = \ e^{6x}.\)

There is no elementary antiderivative for \(\displaystyle y \ = \ e^{2^x}.\)


\(\displaystyle For \ (-\infty, 0), \ y \ = \ e^{2x} \ is \ above \ y \ = \ e^{6x}. \ \\)


The curves intersect at (0, 1). \(\displaystyle And \ y \ = \ e^{6x} \ is \ above \ y \ = \ e^{2x} \ for \ (0, 1) .\)


The area of the region bounded by those curves would be:

\(\displaystyle \lim_{N \to -\infty}\int_{N}^{0}e^{6x}dx \ + \ \int_{0}^{1}e^{2x}dx \ = \\)


\(\displaystyle \frac{1}{6} + \bigg(\frac{1}{2}e^2 - \frac{1}{2} \bigg)\ = \\)


\(\displaystyle \frac{1}{2}e^2 \ - \ \frac{1}{3} \ \approx \ 3.36119\)
 
Ah! So much more sense. I was not getting tthat at all. Excellent translation, lookagain.
 
That answer is incorrect. I appreciate the help though. Any other suggestions would be greatly appreciated as this is a homework assignment and it is due tonight.
 
Sadly, we're still haveng communication difficulty. We don't even know what the problem statement actually is.
 
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