Find a value for "a" such that the average value of the function y = 2^x - 1 on the interval [0,a] is equal to 1.

Answer 1

when x = 0 , y = 2^0 - 1 = 1-1 =
when x = a, y = 2^a - 1

average = (2^a - 1 - 0)/(a-0) = 1
2^a - 1 = a

by inspection a = 1 or a = 0

Answer 2

the average value of f(x) on [a,b] is ∫[a,b] f(x) dx / (b-a) so, we want
∫[0,a] (2^x - 1) dx = 1
(2^a - 1)/ln2 = 1
2^a - 1 = ln2
2^a = 1 + ln2 = ln(2e)
a = ln(2e)/ln2 = log_2(2e) = 1 + ln_2(e) = 1 + 1/ln2

the answer shown above is not the average of f(x), but the average rate of change of f(x) -- the slope of the line joining the two points.

Answer 3

parentheses ????
y = 2^x - 1
or
y = 2^(x - 1)
I will assume it is the first, as you typed it.
integral of b^u du = b^u / ln b
integral of 2^x dx = 2^x/ln 2
integral of 2^x -1 = 2^x/ln 2 - x
evaluate at a
= 2^a/ln 2 - a
evaluate at 0
= 1/ln 2
subtract
(2^a -1)/ln 2 - a = 1
ln 2 = 0.693
(2^a - 1) / 0.693 = 1 + a
2^a - 1 = 0.693 + 0.693 a
2^a - 0.693 a = 1.693
if a = 1
2 - 0.693 =? 1.693
1.307 is not 1.693 diff about .4
if a = 1.1
2.14 - .762 = ? 1.693
1.378 is still not there
if a = 1.5
2.83 - 1.04 =? 1.693
1.79 is much closer to 1.693 but the other side
try a = 1.4 etc

Answer 4

oops - go with Anonymous -- I had a typo in my integral, which threw off the rest.