help with integration (1 Viewer)

[danif]

hey,

how would you do the integration of 3^(2x-1) ?????

what's the general rule for this kind of integration?

thanks for your help

 

[Winston]

lol i haven't touched integration in so many months

is it 3^(2x-2) / (2x - 2) + C

i think that's wrong hahaha

 

[cyrax83]

*tries to do it but has completely forgotton how to integrate*

 

[KeypadSDM]

3^{2x - 1}
= e^{Ln[3^{2x - 1}]}
= e^{(2x - 1)Ln[3]}

Differentiation:
:. d/dx 3^{2x - 1}
= d/dx e^{(2x - 1)Ln[3]}
= 2Ln[3] * e^{(2x - 1)Ln[3]}
= 2Ln[3] * 3^{2x - 1}

Integration:

/
| 3^{2x - 1}dx
/
=
/
|e^{(2x - 1)Ln[3]}dx
/
= e^{(2x - 1)Ln[3]}/(2Ln[3]) + c
= 3^{2x - 1}/(2Ln[3]) + c

 

[Heinz]

This doesnt seem like a 2unit job

 

[RUB!X]

???

 

[danif]

according to the answers, you're right keypadSDM. i'm pretty sure integration is a 2U topic- tho maybe this level is 3U?? if so- my bad.

 

[jesshika]

isn't it an exponential ???

 

[Winston]

ah yeah this does look like 3u, haha its all coming back i remember you cant integrate a function when the indice is an actual function.

 

[Heinz]

Originally posted by Winston
ah yeah this does look like 3u, haha its all coming back i remember you cant integrate a function when the indice is an actual function.

yeah, its probably a harder 2unit application. From what i remember of the 2unit course, these types of questions are usually used in conjunction with the trapazoidal or simpsons rule.

 

[KeypadSDM]

Did anyone actually bother to read what I wrote?

 

[KeypadSDM]

3^{2x - 1}
= e^{Ln[3^{2x - 1}]}
= e^{(2x - 1)Ln[3]}

This can be represented a little easier:

3^{2x - 1} = j^k
j = 3
k = 2x - 1
3 = e^Ln[3]
:. 3^{2x - 1} = (e^Ln[3])^{2x - 1}
= e^{Ln[3](2x - 1)}

 

[Zarathustra]

Originally posted by KeypadSDM
Did anyone actually bother to read what I wrote?

- yes we're all so impressed (sorry if that sounded sarcastic )

btw it is good that 4u dux does help out lowly 2u people

 

[ping pong]

it is 2u!!!!i can see it! lol it took me a while but
soo simple....yet looks so impossible at first!!

its a more complicated version of
x=a to the power of y
and y = log a (x)

if you look carefully where x is 3
and y is (2x-1)


thanx KeypadSDM - u make it look so easy =0P btw how did u type it up so it woz 3^2x - 1 but show the 2x-1 as a power???

 

[Collin]

Originally posted by Heinz
This doesnt seem like a 2unit job

It's 2 unit.