What is #int_1^oo 1/2^x dx#?

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Answer 1 · 2015-11-08T20:13:02

#int_1^oo 1/2^x dx = [-1/(ln(2)2^x) ]_1^oo = 1/(2 ln(2)) ~~ 0.721#

#2^t = e^(ln(2)t)#

So:

#1/(2^x) = e^(-ln(2)x)#

#d/dx (1/2^x) = d/dx (e^(-ln(2)x)) = -ln(2) e^(-ln(2)x) = -ln(2) 1/2^x#

So:

#int 1/(2^x) dx = -1/(ln(2)2^x) + C#

And:

#int_1^oo 1/2^x dx = [-1/(ln(2)2^x) ]_1^oo = 1/(2 ln(2)) ~~ 0.721#