How to integrate a function

[cirthix]

sigh, the integral of e^(2x) is (1/2)e^2x.

for future refrence e^x is its own derivative (and integral).


edit: did you mean the integral of (e^(2x))/x?

 

[Howard]

Oh, that. That was just some **** I tried before, I didn't feel like erasing it for you.

 

[Howard]

Originally posted by: cirthix
sigh, the integral of e^(2x) is (1/2)e^2x.

for future refrence e^x is its own derivative (and integral).


edit: did you mean the integral of (e^(2x))/x?

Isn't that what I said?

 

[Goosemaster]

i jsut worked it out and got as my answer

e^2x - x^2
-------- -------
x ............ .......2


is that what the book said? ( iadded the periods so it would look justified

 

[DrPizza]

Well, duhhh, here's how you integrate it:
http://integrals.wolfram.com/
The Integrator!
🙂 😛

 

[Howard]

Well, that was quick, but I'm not quite sure how to read the output.

ExpIntegralEi[xLog[e]]

 

[DrPizza]

Hmmmm.... the integrator failed me on this one.

 

[Goosemaster]

what does your book say?😛

 

[Goosemaster]

Originally posted by: DrPizza
Hmmmm.... the integrator failed me on this one.

LOL.....



coming from you= x^LOL

 

[DrPizza]

Let me ask this:
Did you copy the problem correctly?
I don't know how anyone is getting a solution, please give a hint at least to the method you're using.

 

[DrPizza]

Never mind... I figured it out... integration by parts. Twice.

 

[TuxDave]

doh...

 

[Goosemaster]

Originally posted by: TuxDave

Originally posted by: Howard
Integration by parts doesn't seem to be working for me. What am I doing wrong?

(I know I should be using different letters, but too late to fix that now)

http://www.rootminus1.com/howard/homework2.gif

int (u*v') = u*v-int(u'*v)

u = e^(2x)
v' = 1/x
v = ln|x|
u' = 2*e^(2x)

int (u*v') = e^(2x)*ln|x|-int(2*e^(2x)*1/x)
=e^(2x)*ln|x|-2*int(u*v')
....
Profit?

Unless I did something wrong.

dude.it's so much easier to use e for differentiation rather than for integration

 

[Howard]

Originally posted by: TuxDave

Originally posted by: Howard
Integration by parts doesn't seem to be working for me. What am I doing wrong?

(I know I should be using different letters, but too late to fix that now)

http://www.rootminus1.com/howard/homework2.gif

int (u*v') = u*v-int(u'*v)

u = e^(2x)
v' = 1/x
v = ln|x|
u' = 2*e^(2x)

int (u*v') = e^(2x)*ln|x|-int(2*e^(2x)*1/x)
=e^(2x)*ln|x|-2*int(u*v')
....
Profit?

Unless I did something wrong.

How many times do I have to integrate by parts?

All this is to solve a linear differential equation, it can't possibly be that difficult.

 

[Howard]

Originally posted by: Goosemaster
i jsut worked it out and got as my answer

e^2x - x^2
-------- -------
x ............ .......2


is that what the book said? ( iadded the periods so it would look justified

Finding the integral is just a small part of the question. Sorry.

 

[TuxDave]

Originally posted by: Howard

Originally posted by: TuxDave

Originally posted by: Howard
Integration by parts doesn't seem to be working for me. What am I doing wrong?

(I know I should be using different letters, but too late to fix that now)

http://www.rootminus1.com/howard/homework2.gif

int (u*v') = u*v-int(u'*v)

u = e^(2x)
v' = 1/x
v = ln|x|
u' = 2*e^(2x)

int (u*v') = e^(2x)*ln|x|-int(2*e^(2x)*1/x)
=e^(2x)*ln|x|-2*int(u*v')
....
Profit?

Unless I did something wrong.

How many times do I have to integrate by parts?

All this is to solve a linear differential equation, it can't possibly be that difficult.

Oops.. nm.. made a mistake

 

[cirthix]

when i integrated by parts twice, i got htat hte integral is equal to itself. hmm

 

[TuxDave]

doh.. mistake

 

[Goosemaster]

http://wisetyro.com/math.jpg

how i did it


you have to finish it though by moving dow the left part where i was done and replacing the u with the letter x😉

edit: when i did IBP I assumed that we all knew that u=x so int(ln(u)) was 1/x

 

[Howard]

Originally posted by: TuxDave

Originally posted by: Howard

Originally posted by: TuxDave

Originally posted by: Howard
Integration by parts doesn't seem to be working for me. What am I doing wrong?

(I know I should be using different letters, but too late to fix that now)

http://www.rootminus1.com/howard/homework2.gif

int (u*v') = u*v-int(u'*v)

u = e^(2x)
v' = 1/x
v = ln|x|
u' = 2*e^(2x)

int (u*v') = e^(2x)*ln|x|-int(2*e^(2x)*1/x)
=e^(2x)*ln|x|-2*int(u*v')
....
Profit?

Unless I did something wrong.

How many times do I have to integrate by parts?

All this is to solve a linear differential equation, it can't possibly be that difficult.

Oops.. nm.. made a mistake

I see what you're getting at, but when I bring it to the other side I lose what I was integrating in the first place.

 

[Howard]

Originally posted by: Goosemaster
http://wisetyro.com/math.jpg

how i did it


you have to finish it though by moving dow the left part where i was done and replacing the u with the letter x😉



edit:


I messed up with the writing but the work is still good

1/u = dv
lnu = v

my bad..problem is still good though

[e^(2u)]/u - int[(lnu)/u] = int(udu)?

 

[Goosemaster]

[e^(2u)]/u - int[(lnu)/u] = int(udu)?

no

1)e^(2u)]/u - int[(lnu)/u]

2)e^(2u)]/u - int(udu)

3) e^(2u)]/u - (u^2)/2 <-----that last part was yet another substitution

I am lazy so I used u again


I lhate writing sh!t over an over again so I jsut keep goign only with what needs work

 

[Howard]

OK if somebody could just try to solve this DE to see if I did something wrong, that would be great.

http://www.rootminus1.com/howard/homework3.gif

First term is x squared times y prime

 

[Howard]

Originally posted by: Goosemaster

[e^(2u)]/u - int[(lnu)/u] = int(udu)?

no

1)e^(2u)]/u - int[(lnu)/u]

2)e^(2u)]/u - int(udu)


I lhate writing sh!t over an over again so I jsut keep goign only with what needs work

How is (lnu)du/u = udu?

You used u = (lnu)/u?

Man, I am so going to fail.

 

[DrPizza]

Originally posted by: DrPizza
Never mind... I figured it out... integration by parts. Twice.

no it isn't either... everything cancelled (I was slowly, but surely typing it in)