integrate 2^x

d/dx(2^x)=(2^x)(ln2)
∫2^x dx
= (2^x)/(ln2)+C
So integrate(1 to -1) 2^x dx
=[2/ln2]-[(-1)/ln2]
=(3/2ln2)

2009-05-15 14:17:14 補充：
根本上就不需要用substitution

integrate(1 to -1) 2^x dx
Sol
Note that d/dx(a^x) = a^x ln a, thus ∫a^x dx = a^x/(ln a)
Therefore,
∫2^x dx [1,-1]
= 2^x/ln 2 [1,-1]
=2 / ln 2 - 1 / (2 ln 2)
= 1/ ln 2 (2-1/2)
= 3/(2 ln 2)

2009-05-15 18:41:50 補充：
從d/dx(a^x)可得上題之答案^^

2009-05-16 22:00:14 補充：
確實不需要substitution.

Let y = 2^x
ln y = x ln 2
so y = e^(x ln 2)
That is to integrate e^(x ln 2)
Let x ln 2 = u
dx = du/ln 2
when x = 1, u = ln 2
when x = -1, u = - ln 2 = ln(1/2)
so the integral becomes
S e^udu/ln 2 = e^u/ln 2 from ln(1/2) to ln 2
= 2/ln 2 - 1/2ln 2
= 3/2 ln 2.
=

Let u = 2x
ln u = x ln 2
du/u = ln 2 dx
∫(-1 to 1) 2^x dx
= ∫(1/2 to 2) u du / u ln 2
= ∫(1/2 to 2) du / ln 2
= (2 - 1/2) ln 2
= 3/2 ln 2

2009-05-16 03:28:03 補充：
「根本上就不需要用 substitution」？

How to get d/dx(2^x)=(2^x)(ln2) or e^lnx = x ?

Does it involve substitution ?

Write (x+1)^2 - (x+1) - 2 = 0, then (x+1 -2)(x+1 + 1) = 0 really avoid substitution ?

We can't prove Pyth thm by cosine law.