A. think if the area as a bunch of strips of width dx. Since the curves meet at (0,0) and (1,1), the area is
∫[0,1] (x - (2^x - 1)) dx
B. following the usual steps,
x = 2^y - 1
2^y = x+1
y = log_2(x+1)
C. f and f-1 are symmetric about the line y=x
D. You want a such that
∫[0,a] (2^x - 1)) dx = 1
(2^a - a)/ln2 = 1
a = -2.108 or 1.421
1. Consider the curve y = f(x) = 2^x - 1.
A. Find the exact area of the region in the first quadrant bounded by the curves y = f(x) = 2^x - 1 and y = x. ("Exact area" means no calculator numbers.)
B. Find the inverse function y = f^-1 (x).
C. Using part A and the notion of symmetry between a function and its inverse, find the exact area of the region in the first quadrant bounded by the curves y = f^-1 (x) and y = x. Explain your reasoning. (Hint: Think "graphically" and little or no math will need to be done!)
D. Find a value for "a" such that the average value of the function f(x) on the interval [0,a] is equal to 1. You may use a calculator here.
5 answers
We have to find the exact area for C, @oobleck.
what do you not understand about symmetry?
B and C are the same area.
B and C are the same area.
I need to write a numerical answer. What should I write?
The answer to C is 2*(3/2-(1/ln(2)), the answer for A. I got 3-(2/ln(2).
1 answer