Expert: Socrates - 11/28/2011

Question
I'm asking about this equation, i hardly do it, my papers are all sketch of it, but still.. no answer popping out

Answer
You want to find

S |x|x^2 dx from -1 to 1  

and

S |x|[sin(x)]^3 / [1+cos(x)] dx from -1 to 1

then add these two values


For the first integral ,


S |x|x^2 dx from -1 to 1 = S |x|x^2 dx from -1 to 0  +  S |x|x^2 dx from 0 to 1  


S |x|x^2 dx from -1 to 0 =  S (-x)x^2 dx from -1 to 0 ( x is negative so |x| = -x)
= S -x^3 dx from -1 to 0 = 1/4

S |x|x^2 dx from 0 to 1 =  S (x)x^2 dx from 0 to 1 = S x^3 dx from 0 to 1 = 1/4


So now we know S |x|x^2 dx from -1 to 1 = 1/4 + 1/4 = 1/2


Next we need

S |x|[sin(x)]^3 / [1+cos(x)] dx from -1 to 1

The easy way to do this is just to observe that the inegrand is an odd function , that means

|-x|[sin(-x)]^3 / [1+cos(-x)] = (-1)|x|[sin(x)]^3 / [1+cos(x)]

The integral of an odd function on an interval symmetric around zero [-a,a] must be zero.


If you don't know this fact then use a change of variable to find

S |x|[sin(x)]^3 / [1+cos(x)] dx from -1 to 1

Let x = -y , then dx/dy = -1 and dx = -dy . The y limits are then from 1 to -1

S |x|[sin(x)]^3 / [1+cos(x)] dx from -1 to 1 = S |-y|[sin(-y)]^3 / [1+cos(-y)] -dy from 1 to -1

= - S |y|[sin(y)]^3 / [1+cos(y)] dy from -1 to 1

Letting I = S |x|[sin(x)]^3 / [1+cos(x)] dx from -1 to 1 , this proves that I = -I , and thus, I = 0

We now have  

S (|x|x^2 + |x|[sin(x)]^3 / [1+cos(x)])  dx from -1 to 1 = 1/2 + 0 = 1/2