## Thefaceless Group Title Calculus 2 problem. Help please. y=e^x, y=e^2x, x=ln2, x=0 Find the area enclosed by the curves. Show steps please. one year ago one year ago

1. dumbcow Group Title

$A = \int\limits_{0}^{\ln(2)} (e^{2x} - e^{x}) dx$

2. Thefaceless Group Title

Right, but what is the actual area. That's where I am having trouble. thanks.

3. Thefaceless Group Title

The answer is supposed to be 1/2, but I am not getting that.

4. zepdrix Group Title

Were you able to get through the integration step? Before plugging in the limits.

5. zepdrix Group Title

Were you able to get this? :) $\large \frac{1}{2}e^{2x}-e^x$

6. Thefaceless Group Title

Yes, I got that part, but I did the algebra wrong, looking at it now. I got the answer now. I have a quick question, however. How do you know which function to use first in the integration process?

7. Thefaceless Group Title

And is there anyway to give more than one medal?

8. zepdrix Group Title

No, just give to cow c: hehe

9. zepdrix Group Title

You mean which one is being subtracted when you setup the integrand? Or do you mean which limit you're subtracting when you evaluate the integral (plugging in).

10. Thefaceless Group Title

I guess both. I thought I was okay on the integral, but I am questioning myself on it now.

11. zepdrix Group Title

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12. zepdrix Group Title

We subtract the lower from the upper function.

13. zepdrix Group Title

So in that example, the area between the curves would be f(x)-g(x).

14. zepdrix Group Title

Confusing picture? :D lol so many arrows, maybe that was silly of me :)

15. Thefaceless Group Title

Lol, I understood it, so it did its job! Basically, though, we are just sticking to the convention of biggest minus smallest, right? Same with the intervals?

16. zepdrix Group Title

Yes, same with the limits. If we did it the other way around, we would end up with the same answer, but negative. And negative area doesn't make much sense ^^

17. Thefaceless Group Title

I see. So what if you had y=cos 2x, x=pi/ 4, x=pi/2 ? Which interval would be on top?

18. Thefaceless Group Title

Is it the one that is biggest relative to the function?

19. zepdrix Group Title

Bah! These ones always confuse me because they didn't give us a second boundary for $$y$$. I think what happens is, our other boundary is just the x-axis $$y=0$$. And in such a case, we're not talking about area between curves. So in this case we can end up with negative area. $\huge \int\limits_{x=\pi/4}^{\pi/2} \cos2x-0\; dx$

20. zepdrix Group Title

Lemme verify that on wolfram real quick to make sure i'm not talking out my butt :P

21. zepdrix Group Title

Yah that seems to be accurate c:

22. Thefaceless Group Title

Hm, when you do it that way, the answer is -1/2. The answer in the back of my book is 1/2.

23. zepdrix Group Title

Really? Hmm wolfram is giving me -1/2 as well... Hmm

24. Thefaceless Group Title

Ohhhhh, I think I know why.

25. Thefaceless Group Title

That part of the curve is under the y axis, but since we can't have a negative area, we abs value the function. I think.. lol

26. Thefaceless Group Title

err the x axis*

27. zepdrix Group Title

Well this is the way I learned in back in calc 2 :D maybe I'm not remembering correctly lol. If we're looking for area between curves, it's always positive because we're thinking of an amount of space between them. But when we're looking for area under a curve, it can be negative. You might wanna ask your teacher about that. Because clearly I'm a little rusty XDD

28. Thefaceless Group Title

I would, but this hw is due first thing in the morning lol. Thanks for your help, though. It cleared up a lot for me! :)

29. zepdrix Group Title

ah i see c: