Printable View
The integral of e^x is e^x. That's on every maths site. But what they don't tell you is why. Is it because the integral of e^x = ( 1/(differential(x)) ).e^x? I want to integrate e^ something slightly more complicated than x so I need to know this. I can't believe it's not on the Internet anywhere.
Eh I have the proof for it somewhere in one of many binders. And no, I'm not going to go dig through it for you. I could make a 6 inch thick book with my calc binders alone.
What is the "slightly more complicated than x"?
look at the bottom of
http://www.jtaylor1142001.net/calcja...ts/CLogExp.htm
e is a "magical" number though. The Integral of of 1/x evaluated from 1 to e is 1. I believe that is where it came from (but again, I don't remember, so don't quote me on that).
int(a^x, dx) = (a^x)/ln(x) + C
that's the rule, so replacing a with e the denominator is ln(e), which is 1 by definition
e^x is defined as the inverse of ln(x). it all stems from that.
Informal proof:
Take y=ln x.
ln x= int(1/x dx from 1->x).
dy/dx = dy/dx * (integral (1/t)dt from 1->x) = 1/x by fundamental theorem of calculus.
so the derivative of ln x is 1/x by FTOC. .
SO THEN we can go ahead and define e = inverse of ln 1.
e^x = inverse of ln x.
so take y = e^x
ln y = x
differentiate both sides with respect to x:
1/y * dy/dx = 1
so dy/dx = y = e^x.
w00t.
You cant do anything with e^x without ln x.
Another good way to think about it is in terms of the Maclaurin expansion for e^x.
e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! +....
There is a theorem that says we can differentiate the Maclaurin/Taylor polynomial term by term and it approaches the actual value, so:
d/dx = 0 + 1 + x + x^2/2 + x^3/3! + .....
(just take the derivative of each term).
so... de^x/dx = e^x. I think this is the way I proved it in my analysis class, albeit formally.
Don't you mean int(a^x, dx) - (a^x)/ln(a) + C? So in that case the power to which e is raised would have to bearing on the integral?Quote:
Originally Posted by stormy
It might be easier if I give the actual integral. Give the integral of x.e^(-x^2/2), that's x times e to the power of minus x-squared-over-two. The answer is -(e^(-x^2/2)), that's minus e to the power of minus x-squared-over-two. I was going to use integration by parts (a technique I only just discovered!) but I need to derive the integrals of the parts.
Yea you need to use integration by parts to solve it. As for the e^(-x^2/2) thing, wouldnt you use substitution? e^u where u=x^2/2 blah blah? Thats at least what jumps out at me.
Depending on how far you go in math, youll learn to hate, and then love, integration by parts. Hate it because they make you do a million of them in your basic Calc classes, love it because your later classes will use it sparingly but help you out tremendously when they do. ^_^
You can just use u-sub the whole way u = the exponent, du = -x
so you have -(e^u) du, no IBP, which will give you the right answer
Unless I'm being a total idiot, LAST WEEK OF MATH EVER so screw that
I graduated at the end of March, I was so happy to get done with Math... but you know, now I kinda miss it. Its weird.Quote:
Unless I'm being a total idiot, LAST WEEK OF MATH EVER so screw that