maths q please (1 Viewer)

[hatty]

how do i integrate

3^(2x-1)


indefinite integral

regards

 

[KeypadSDM]

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

d/dx 3^(2x-1)
= 2Ln[3] * e^((2x-1)Ln[3])
= 2Ln[3] * 3^(2x-1)

EDIT: Whoops, you said integrate. Gimme a sec.

/
|3^(2x-1)dx
/
= Ie^((2x-1)Ln[3])dx

Note that the integral is of the form e^f(x) where f(x) is linear

f(x) = 2Ln[3] * x - Ln[3]
f'(x) = 2Ln[3]

:. Ie^((2x-1)Ln[3])dx
= [e^((2x-1)Ln[3])]/(2Ln[3]) + c
=3^(2x-1)/(2Ln[3]) + c

 

[hatty]

i dont understand
the font is 2 hard 2 read
thanks n e way key

 

[KeypadSDM]

Originally posted by hatty
i dont understand
the font is 2 hard 2 read
thanks n e way key

Sorry

Hold control and wind the mousewheel wither way and it gets bigger and smaller. Or copy it into word.

It's the same as differentiation, but instead of multiplying, divide by f'(x) (Only when f(x) is linear)

 

[Grey Council]

way outta my league.

Is this a hard 4u integration question, Keypad? Or is this typical?

 

[ND]

This is actually a 3u question, and is not hard at all. All you have to do is convert it to e^something, then integrate as you normally would.

 

[KeypadSDM]

It only looks hard because, well, I'm typing it. On paper it looks ALOT simpler.