Simple integration to find area

[Zman]

Homework Statement

I wish to find the area under the curve y = 1/2^x between x=0 and x=1 but get an answer that is half the expected answer.

Homework Equations

Integrate y = 1/2^x to get -1/(2^x ln2) + Const

This integration result was confirmed on Wolfram

Slot in the range x = 0 to x = 1

Area = 1/ln2 (1 - 0.5) = ~0.34

The Attempt at a Solution

But when I look at the curve of y = 1/2^x

I can see that the area under the curve between x=0 and x = 1 is double

i.e. ~0.68The range of the curve y = 1/2^x that I am interested in starts at point (0,1) and goes to point (1, 0.5)

A rough visual calculation of this area is made up of the lower rectangle 1 x 0.5 added to the area of the upper triangle (1x0.5)/2 = 0.25 which gives an approximate answer of 0.75I seem to be missing a factor of 2 as regards the integral and am at a loss.

 

[haruspex]

1/ln2 (1 - 0.5) = ~0.34

You seem to have calculated ln(2)/2, not 1/(2ln(2)).

 

[Zman]

Thank you so much, this grey head is so greatful.