Integration problem (1 Viewer)

[aud]

I'm really not getting this section on Logs and Exps... can someone tell me how I should do this question?

Find y, given that y' = 2^-x, and y = 1/(2log2) when x = 1

I've spent so long on it, and I know that when someone shows me it's going to be so obivious...

Thanks...

 

[Xayma]

Its an power one so use first principles to differentiate 2^-x and then apply the reverse to the integration and then you have a constant so use it to make it equal to that other one when x=1

 

[~*HSC 4 life*~]

huh? oh dear im so confused...

why would you differentiate 2^-x?

i thought you woudl try and integrate it perhaps...except i dont know how to integrate thal lol :/

 

[Affinity]

y' = 2^-x

y = integrate 2^-x dx

y = integrate e^[-x ( ln(2) )] dx

y = - [ 1 / ln(2) ] * e^[ - ln(2) x ] + C

then you substitute your initial value to find C.

 

[SamD]

I haven't done this stuff for ages, but here goes....

First get y' into a form we can differentiate...
y'=2^-x
lny'=ln(2^-x)
lny'=-x ln2
y'=e^(-xln2)

Differentiate this sucker, seems strange but it helps...
y''=-ln2.e^(-xln2)
= -ln2.2^(-x)

This means that...
Integral -ln2.2^(-x) dx= 2^(-x) + c
-ln2 Integral 2^(-x) dx= 2^(-x) +c

The integral bit is really y... so...
-y ln2 = 2^(-x) +c

Note -c/ln2 is just a constant, so we'll just keep calling it c.
y = -2^(-x) / ln2 +c

Now find c, so substutute in the point from the question...
1/2ln2 = -2^(-1)/ln2 +c
1/2ln2 = -1/2ln2 +c
c=2/2n2
c=1/ln2

Therefore.
y= -2^(-x)/ln2 + 1/ln2
y=(1 - 2^(-x))/ln2

I think this is right...

Sam

 

[KeypadSDM]

Quickly:

y' = 2^-x
= e^(-xLn[2])
:. y = (e^(-xLn[2]))/(-Ln[2]) + c
= -(2^-x)/(Ln[2]) + c

y = 1/(2Ln[2]), x = 1

:. 1/(2Ln[2]) = -1/(2Ln[2]) + c
c = 1/(Ln[2])

:. y = -(2^-x)/(Ln[2]) + 1/(Ln[2])
y = (1 - 2^-x)/(Ln[2])

 

[SamD]

Oops... Thanks Keypad... I got c wrong... Will edit my post...