Expert: Alon Mandes - 9/21/2008

Question
I am trying to integrate  (e^x   - 1 )^2  with definite limits  1 (at top ) and -1 (at bottom).  I know answer is
1/2e^2  (e^4  -  4e^3  +  4e^2  + 4e  -  1 )
I have tried for a while but can#t get the e^4 or e^3 terms
Would appreciate an explanation

Answer
Ok Carole, the first think we will do is deal with the power :
We know that (a+b)^2=a^2-2ab+b^2, so (e^x-1)^2=(e^x)^2-2e^x+1.
We also know that (a^m)^n=a^(mn), so (e^x)^2=e^2x. Now our integral
becomes : ∫[e^2x-2e^x+1]dx=∫[e^2x]dx-2∫[e^x]dx+∫dx {from -1 to 1}
Ok, Let's remember : the derivative of e^2x is 2e^2x, that means
the derivative of (e^2x)/2 is e^2x. & we also remember that the derivative of e^x is e^x & we know that ∫dx=x. Thus, we may now
calculate the integral : Let's first calculate every compound aside :
1. ∫[e^2x]dx=(e^2x)/2 {x goes from -1 to 1} =
  =(1/2)e^(2*1)-(1/2)e^(2*(-1))
  =(1/2)e^2-(1/2)e^(-2)
2. ∫[e^x]dx=e^x {x goes from -1 to 1} = e^(-1)-e^1=(1/e)-e
3. ∫dx=x {from -1 to 1} 1--1=1+1=2
Allright, now let's combine :
∫[(e^x)-1)^2]dx=(1/2)e^2-(1/2)e^(-2)-2[(1/e)-e]+2.
Hence, the answer is (1/2)e^2-(1/2)e^(-2)-(2/e)-2e+2
Let's try to make some cosmetic changes, let's take out (1/2e^2)
as a common multiplier, this gives us :
(1/2)e^2-(1/2)e^(-2)-(2/e)-2e+2=(1/2e^2)*[e^4-1-4e-4e^3+4e^2]
If you open the brackets you will get the same form on the left.

Alon.