needs some background knowledge >< (1 Viewer)

[fullonoob]

differentiate xe^x and hence integrate xe^x
d/dx = e^x(1+x)
now how do you integrate it without integration by parts?

 

[Pwnage101]

 

[fullonoob]

oh damn thx..whats this called when you chuck it on the other side? does it have a name or is it just normal integration?

 

[fullonoob]

given that e^logx = x, what is 2^logx?
e^logxlog2??
hence integrate 2^logx
dno how to convert it properly...

 

[hscishard]

Isn't C just +2?

The asymptote for -3^-x is y=0

Shift it up two spaces and the asymptote becomes y=2.

 

[hscishard]

Eh wtf you just changed the question. Lol

I got 2^logx times log2

But it doesn't make sense, integrate 2^logx.
If you use the formula straight away..
2^logx / log2 + C?

 

[fullonoob]

Eh wtf you just changed the question. Lol

I got 2^logx times log2

But it doesn't make sense, integrate 2^logx.

haha we're both wrong, sub in calc
help anyone?

 

[hscishard]

Then it has to be log2(2^logx)


Eeep. Nvm, logx is variable.

 

[Pwnage101]

 

[fullonoob]

lol i got same answer as you
got confused as the question says integrate between 1 and 0
but the ln0 part gives "maths error (cant remember what its called xD)" does not exist??
then i thought of anything times 0 gives zero so yeah
thanks

 

[fullonoob]

heres another question that you will have fun with mod
i haven't really tried this question, but looking at it, i know it will take me a while to figure out so im gna go do something else instead LOL
thanks for your hard work once again ^^

 

[Pwnage101]

I'm a Chemistry/Physics mod, but always happy to help on the maths forums. As for your last question, it is rather trivial (and indeed the answer shows you that), so i'll leave that to you.

 

[fullonoob]

i dno what you do now actually xP
gave it a go
just thought of splitting the fraction, try simplify then tired ><

 

[super.muppy]

(e^x+1)/(e^x/2+e^-x/2) = (e^x+1)/(root e^x + 1/root e^x) = (e^x+1)/((e^x+1)/root e^x) = (root e^x)(e^x+1)/(e^x+1) = root e^x = e^x/2

integrate that = 2e^x/2 + C

now teach me how to use equation editor

 

[Pwnage101]

jetblack (Mathematics moderator) has a very useful guide stickied to these forums on how to use the LaTeX equation editor.

 

[fullonoob]

may god jet black answer my question please

 

[randomnessss]

 

[fullonoob]

wow thanks a lot didnt think of changing the indices
i have another one, it made me rage quit cos i used so many ways and ended up nowhere....



please help

 

[Drongoski]

very easy!

 

[Drongoski]

very easy!

if you are not comfortable with this approach,

d(sqrt x)/dx = 0.5/sqrt x ==> dx/sqrt x = 2 d(sqrt x)