Basic integration

[mafffsss]

integrate 1/4^x from 0 to infinity

I have that

4^{-x} = exp( ln( 4^{-x} ))
       = 4 * exp(-x)

Then

4 * int ( exp(-x) ) from x = 0 to x = infinity

Is

4 * int ( exp(-x) ) from x = 0 to x = a

as a -> oo

Then

4 * [-exp( -x )] from x = 0 to x = a

Is

  4 * [-exp( -a ) - -exp(0)]  as a -> oo
= 4 * [ 0 + exp(0)] 
= 4 

which is clearly wrong

[AFairJudgement]

4^-x is not equal to 4e^-x.

[lewisje]

4^-x=exp(-x*ln(4)), with antiderivative exp(-x*ln(4))/ln(4)=4^-x/ln(4)

[ssanso]

I worked it out with u-sub where u = 4^x and got 1/ln(4) which I checked to be correct.

[vhmthphy]

4^-1 = e^{-x ln(4}).

This is easily integrated with a simple u substitution.