Calculus

Someone please explain to me why the integral of e^u equals e^u. In my head it makes sense but i cant prove it.

[eqn]\int e^x[/eqn] hehe

what is the derivative of e^x??

because [math]{d\over dx}e^u=e^u[/math]

duh

69

It is the function whose derivative is itself
y'=y

Nobody knows, it's a mystery.

Because it just is. Prove it to yourself using Riemann sums.

taylor expand and derive u will see

the integral is the accumulation of the function; hence why you can use the definite integral to find the area under a curve

ration it like this; every point on the graph of e^x represents BOTH the slope of the tangent line of the function as well as the accumulation (sum of infinitesimals) up to that x value. hence why it is its own derivative and integral

take its derivative

if that doesn't make sense use the limit definition of a derivative

if you don't understand why the integral is the antiderivative, do a riemann sum

integrate the taylor series and it'll make more sense

It's its own grandfather m80

draw a graph of e^x and you'll quickly find out you need e^x of lead to colour under the graph.

>dx
>u

Math is more pointless than videogames.

you dont need to prove it faggot. youre in calc 2

How are you defining [math] \exp x [/math]?
Using this?
[eqn] \ln x = \int^1_0 \frac{1}{x} [/eqn]
Or this?
[eqn] \exp x = \sum^\infty_{i=0} \frac{x^i}{x!} [/eqn]

*pardon any mistakes. I just woke up and can't be arsed to look up my taylor series.

Remember that e is a constant number. exp(1) is ~2.718

Also, you're integrating in respect to x. e is not a variable.

>{d\over dx}
>this actually works
fugg

...

Well done op, nice bait.

I hope you're enjoying that electronic device that's doing billions of pointless things per second.

You spent more time TeXing than thinking, didn't you?

Idk how to use the math plugin, I'm dumb.
Axioms:
d/dx ln(x) = 1/x
d/dx x=1

d/dx ln(e^x) = d/dx x=1
d/dx ln(e^x) = d/du ln(u) d/dx e^x //u = e^x
= 1/u d/dx e^x
= 1/e^x d/dx e^x
= 1
Q.E.D

Could copy my last year notes with comments, but too lazy brah

Isn't it basically defined that way?

Yes and no. It more so happens that the L(x) e^x, dy/dx e^x and the integral of e^x dx are all e^x, as such it is incredibly useful for hyperbolic series.

because e is the transcendental connective tissue between functions of different powers.

You stupid fat fuck

cease and desist

it doesnt.

[math]\int e^x dx = e^x + c[/math]

What OP wrote is true because what you wrote is true. OP isn't wrong.

OP gave a particular solution.
i personally dont think that is sufficient.

> the integral of a function
> no precision of variables type
Assuming it's the real defined and valued function
> no bounds
Stopped thinking about it. Reformulate your question.

>intelligent af
>too lazy bruh
literally every Veeky Forums user

e^x = 1+x+x^2/2!+x^3/3!+x^4/4! ...

int e^x dx = 1+x+x^2/2!+x^3/3!+x^4/4! ... + c' || c' = c-1

int e^x dx = e^x + c'

Did not bother to format in latex because it is this easy to show. The constant is c' for absolute dummies who can't see "where the 1 goes".

ITT: op trolls people into writing 'sex' a bunch of times

[math]\int_{X}^{} a_uf_udu[/math]

Please write the whole sentence user:
[math] \int e^{x} = f(u)^{n} [/math]
.
.
Proof on request.

Prove it.

Let [math]f(a)=(e^x)^{1/n}[/math].
.
[math]\lim_{t \rightarrow \infty} \left( f(a) - \frac{1}{f(t)} \right)=\frac{d}{dx}f(u) \Leftrightarrow \lim_{t \rightarrow \infty} \left( f(a) - \frac{1}{\infty} \right) [/math]
[math] = \frac{d}{dx}f(u) \Leftrightarrow \lim_{t \rightarrow \infty} \left( f(a) - 0 \right) [/math]
[math]= \frac{d}{dx}f(u).[/math]
.
That yields:
.
[math](e^x)^{1/n} = \frac{d}{dx}f(u) \Leftrightarrow e^x[/math]
[math] = \frac{d}{dx}f(u)^n \Rightarrow \int e^x [/math]
[math] = \int \frac{d}{dx}f(u)^n.[/math]
.
Let's conclude:
.
[math]\int e^x = f(u)^n[/math].

Better display.

Let [math]f(a)=(e^x)^{1/n}[/math].
.
[math]\lim_{t \rightarrow \infty} \left( f(a) - \frac{1}{f(t)} \right)=\frac{d}{dx}f(u) [/math]
[math]\Leftrightarrow \lim_{t \rightarrow \infty} \left( f(a) - \frac{1}{\infty} \right) = \frac{d}{dx}f(u) [/math]
[math]\Leftrightarrow \lim_{t \rightarrow \infty} \left( f(a) - 0 \right) = \frac{d}{dx}f(u).[/math]
.
That yields:
.
[math](e^x)^{1/n} = \frac{d}{dx}f(u) \Leftrightarrow e^x = \frac{d}{dx}f(u)^n [/math]
[math]\Rightarrow \int e^x = \int \frac{d}{dx}f(u)^n.[/math]
.
Let's conclude:
.
[math]\int e^x = f(u)^n[/math].

what a complete bullshit proof.

Did you get this is a meme?

Have you even understood there is a sentence involved in the conclusion of this "proof"?

Because you seem to be such a retard, I am giving myself the right to call you a faggot.

yes ofcourse i get what you are saying retard.

sex is fun hahaha what a nice meme.

still you could have put in effort to make an actual proof. not just that bogus you wrote down.
kys

dumb cs fag cant read math..
heres something in your language.. haha
you == fail && me.KEK() ? die whore