Expert: Alon Mandes - 10/19/2010

Question
Evaluate the given definite integral. a=-1 ; b=1
[(X^2 + 6X -2)]^2 / 9 dx. Give you please all details for each step of the problem solution. I am a little slow on picture up the steps.
Thanks!

Answer
Hello Alvin,
Lets do the solution elaborated step by step progress :

1st, let's examine the form inside the integral : (χ²+6χ-2)² .
Let's perform the power : We know that (a+b+c)²=a²+b²+c²+2ab+2bc+2ac
Thus, (χ²+6χ-2)²=(χ²)²+36χ²+4+2χ²(6χ)+2(6χ)(-2)+2χ²(-2)=
               =χ^4+36χ²+4+12χ³-12χ-4χ²
               =χ^4+12χ³+32χ²-12χ+4

Thus our integral becomes :

+1
∫  (1/9)*[χ^4+12χ³+32χ²-12χ+4] dχ = ?
-1

The integral is a liear operator. That means : ∫a[f(χ)+g(χ)] dx =
a∫f(χ)dχ + a∫g(χ)dχ .

This will lead to :  

+1
∫  (1/9)*[χ^4+12χ³+32χ²-12χ+4] dχ =
-1

(1/9) [ ∫χ^4 dχ + ∫12χ³ dχ + ∫32χ² dχ - ∫12χ dχ + ∫4dχ ] .
Let's calculate each term alone, AND then we will sum the result .
Let's keep in mind the rule : ∫χ^m dχ = (1/m+1)χ^(m+1) .
So,
∫χ^4  dχ = (1/5)χ^5. If we substitute the limits we get :
          (1/5)(1)^5 - (1/5)(-1)^5 = 2/5 .
∫12χ³ dχ = 3χ^4 . If we substitute the limits we get :
          3(1)^4 - 3(-1)^4 = 0
∫32χ² dχ = (32/3)χ³ . If we substitute the limits we get :
          (32/3)(1)^3 - (32/3)(-1)^3 = 64/3
∫12χ  dχ = 6χ² . If we substitute the limits we get :
          6(1)^2 - 6(-1)^2 = 0  
∫4    dχ = 4χ . If we substitute the limits we get :
          4(1) - 4(-1) = 8

Thus , the result will be : (1/9)[2/5 + 64/3 + 8] = 3.3

Alon.