Integrating and differentiating the number e.

[theCandyman]

In calculus, my work has recently involved integrating and differentiating the numer e, of which I am very unsure of how to do. I set up some examples for myself to try to figure out, could anyone tell me if they are correct? Please correct me if I am wrong, or tell me where I have made a mistake.

1) [itex]\int_{}^{} edx = e[/itex] Simple, or it should be. This is how I interpet it from what I have read. The text always makes basics sound confusing to me.

2) [itex]\int_{}^{} e^xdx = e^x[/itex] The same as above because [itex] ln(e) = 1[/itex].

3) [itex]\int_{}^{} e^{2x}dx = \frac{1}{2x+1}e^{2x+1}[/itex] No clue here, to me it looks completely wrong.

4)[itex]\frac{d}{dx} e^{2x} = \frac{1}{2}e^{2x}[/itex] Chain rule applies here, correct? Or is it just [itex]\frac{d}{dx} e^{2x} = e^{2x}[/itex]?


Also, I did a search for another thread that would explain this but did not find one. If one of you reading this could post the link if there is another thread discussing it, or if you know of a really good web page where all of this is explained in a simple, straight-foward way, I would appreciate it very much.

I apologise if this belongs in the homework section, but this is not exactly homework.

 

[Justin Lazear]

[tex]\int e dx = e \int 1 dx = e x[/tex]

e is just a constant. Don't forget to treat it like one.

2 is correct, but I don't understand your reasoning. The derivative of e^x is e^x, so the antiderivative of e^x should also be e^x. I don't see where logarithms come in, or why log(e) = 1 is useful.

3 is wrong. Use the substitution u = 2x and du/2 = dx to make an integral you know how to deal with. Also, you can be sure your solution is not correct by differentiating your answer. You should get back the integrand. Also keep in mind that the equation you mistakenly used in this case is for x raised to a constant power, not a constant raised to the x power. That result is different.

4 is fine.

--J

 

[Hurkyl]

(4)'s wrong -- he should've multiplied by two.


Technically, they're all wrong. e.g.

[tex]
\int e^x \, dx = e^x + C
[/tex]

Where C denotes an arbitrary constant. Don't forget that indefinite integrals have many solutions, one for each choice of constant.

And for (4), I think he meant to write:

[tex]
\frac{d}{dx} e^x
[/tex]

 

[Justin Lazear]

Yup.

--J

(extra characters added here.)

 

[HallsofIvy]

Actually, he's doing everything wrong:

[itex]\frac{dy}{dx}e^{2x}[/itex] doesn't even make sense. It should be either

[itex]\frac{de^{2x}}{dx}= 2e^{2x}[/itex] or [itex]\frac{dy}{dx}= 2e^{2x}[itex] as long as [itex]y= e^{2x}[/itex].

Looks to me like "theCandyman" has a serious problem with the basic concepts of functions and function notation.

 

[theCandyman]

[itex]\frac{dy}{dx}= 2e^{2x}[itex] as long as [itex]y= e^{2x}[/itex].

What does this mean? Sorry about the notation, and thank you for pointing it out. I am just too used to writing that whenever I do differentiation.

Thank you, Hurkyl. I was so concentrated on trying to find what everything was that I forgot the constant. I wrote four correctly, is what HallsofIvy wrote the answer then?

Justin Lazear, [itex]\int_{}^{}k^x = \frac{k^x}{ln(k)}[/itex], this is where I got [itex]ln(e) = 1[/itex].

 

[learningphysics]

For part 4, is this what you wanted:

[tex]\frac{d}{dx} e^{2x}[/tex]

Or in other words the derivative of [tex]e^{2x}[/tex]?

If so then the answer is [tex]2e^{2x}[/tex]

But
[tex]\frac{dy}{dx} e^{2x}[/tex] does not make sense.

 

[dextercioby]

BTW,noone mentioned anything about the title of the thread.Quite interesting,i must say...
Please compute the derivative and the antiderivative for the function:
[tex] y(x)=e [/tex]

Daniel.

 

[theCandyman]

As Justin Lazear said, "e is just a constant. Don't forget to treat it like one." He already integrated e, the derivative is 0.

 

[digink]

BTW,noone mentioned anything about the title of the thread.Quite interesting,i must say...
Please compute the derivative and the antiderivative for the function:
[tex] y(x)=e [/tex]

Daniel.

Isn't e one of the only thing in calculus who's derivate is itself? I don't know if your asking a question or answering one, so ill answer

 

[learningphysics]

Isn't e one of the only thing in calculus who's derivate is itself? I don't know if your asking a question or answering one, so ill answer

e is just a constant. so its derivative is 0.

But the derivative of [tex]e^x[/tex] is itself.

 

[digink]

e is just a constant. so its derivative is 0.

But the derivative of [tex]e^x[/tex] is itself.

Sorry I got confused, I thought I saw e^x, thanks for the clarification