Center of Mass

we know the torque of a body is $ \tau =F\times r$, f is force and  r is vector to the origin.
imagine a horizontal strip with 3 point on it.$\left\{ x_ 1,x_ 2,x_3 \right\} $ with masses :$\left\{ m_ 1,m_ 2,m_3 \right\} $ and a fulcrum $\overset { \_  }{ x } $ on the x-axis origin.
the system might balance, or might not, it depends on how masses are arranged.

each point $m_k$ exert a force toward the ground,according to the gravity force. the force is $m_{ k }.g$;
and the d vector in here is $x_k$. so torque is $\tau =m_{ k }.g\quad \times \quad x_{ k }$;
we know how to balance the system, we should place the fulcrum where the sum of torques would be zero.

$$
F=m.g\\ r=x-\overset { \_  }{ x } \\ \\ \sum { \tau _{ k } } =\sum { F_{ k }\times r_{ k } } =0\\ \sum { m_{ k }.g.(x_{ k }-\overset { \_  }{ x } ) } =0\\ g\sum { m_{ k }.(x_{ k }-\overset { \_  }{ x } ) } =0\\ \sum { m_{ k }.x_{ k }-m_{ k }\overset { \_  }{ x }  } =0\\ \sum { m_{ k }.x_{ k } } -\sum { m_{ k }\overset { \_  }{ x }  } =0\\ \sum { m_{ k }.x_{ k } } =\sum { m_{ k }\overset { \_  }{ x }  } \\ \overset { \_  }{ x } =\frac { \sum { m_{ k }.x_{ k } }  }{ \sum { m_{ k } }  } \\
$$

we call the numerator "system moment" and denominator "system mass'.
imagine a strip from x=a to x=b, we can divide the stripe to small piece of mass $\Delta m$
then we can say the $\overset { \_  }{ x } $ is approximately equal :

$$
\overset { \_  }{ x } \approx \frac { \sum { \Delta m_{ k }.x_{ k } }  }{ \sum { m_{ k } }  }
$$
if the density of strip at $x_k$  is $\sigma (x_k)$ then we can write :
$$
\Delta m_{ k }=\sigma (x_{ k }).\Delta x_{ k }\\ \\ \overset { \_  }{ x } \approx \frac { \sum { x_{ k }.\sigma (x_{ k })\Delta x_{ k } }  }{ \sum { \sigma (x_{ k })\Delta x_{ k } }  } \\
$$

the numerator and denominator are a Riemann sum of density function. we can write it as integral :
$$
\overset { \_  }{ x } =\frac { \int _{ a }^{ b }{ x\sigma (x)dx }  }{ \int _{ a }^{ b }{ \sigma (x)dx }  } $$

limit of Riemann sum

consider f(x)=x and partitions P , we want the area under the equation f:

$$
f(x)=x\\ P=\left\{ \frac { b }{ n } ,2\frac { b }{ n } ,3\frac { b }{ n } ,4\frac { b }{ n } ...n\frac { b }{ n }  \right\} \quad a<b,\quad a=0\\ \sum _{ k=1 }^{ n }{ f(x_{ k }).\Delta  } x_{ k }=\sum _{ k=1 }^{ n }{ x_{ k }.\Delta  } x\\ =\sum _{ k=1 }^{ n }{ k } \frac { b }{ n } .\frac { b }{ n } =\sum _{ k=1 }^{ n }{ k } \frac { b^{ 2 } }{ n^{ 2 } } \\ =\frac { b^{ 2 } }{ n^{ 2 } } \sum _{ k=1 }^{ n }{ k } \\ =\frac { b^{ 2 } }{ n^{ 2 } } .\frac { n(n+1) }{ 2 } \\ =\frac { b^{ 2 } }{ n } .\frac { (n+1) }{ 2 } \quad \\ =\frac { b^{ 2 } }{ 2 } (1+\frac { 1 }{ n } )
$$

as norms |P| are equal , then :

$$
\lim _{ n\rightarrow \infty  }{ \frac { b^{ 2 } }{ 2 } (1+\frac { 1 }{ n } ) } \\ =\lim _{ n\rightarrow \infty  }{ \frac { b^{ 2 } }{ 2 } .\lim _{ n\rightarrow \infty  }{ (1+\frac { 1 }{ n } ) }  } \\ =\int _{ 0 }^{ a }{ x.dx } =\frac { b^{ 2 } }{ 2 }
$$

for interval a,b :

$$
\int _{ a }^{ b }{ f(x).dx } =\int _{ a }^{ 0 }{ f(x).dx } +\int _{ 0 }^{ b }{ f(x).dx } \\ =-\int _{ 0 }^{ a }{ f(x).dx } +\int _{ 0 }^{ b }{ f(x).dx } \\ =-\int _{ 0 }^{ a }{ x.dx } +\int _{ 0 }^{ b }{ x.dx } \\ =-\frac { a^{ 2 } }{ 2 } +\frac { b^{ 2 } }{ 2 } \\ \\ \int _{ a }^{ b }{ x.dx } =\frac { b^{ 2 } }{ 2 } -\frac { a^{ 2 } }{ 2 } \quad ,\quad a<b
$$

The Substitution Rule

$
\int { f(g(x))dx=\int { \frac { d }{ dx }  }  } F(g(x))dx\quad \quad u=g(x),\quad y\quad =F(u)\\ \frac { dy }{ du } =F'(u)=f(u)\quad ,\quad \frac { du }{ dx } =g'(x)\\ \int { \frac { dy }{ du }  } .\frac { du }{ dx } .dx=\int { \frac { dy }{ du } .du= } \int { F'(u) } du=\int { f(u)du }
$

apply to function :

$
\int { x^{ 2 }sin(x^{ 3 }) } dx\quad ,\quad u=x^{ 3 }\\ =\frac { 1 }{ 3 } \int { sin(x^{ 3 }) } 3x^{ 2 }.dx\quad ,\quad du/dx=3x^{ 2 }\\ =\frac { 1 }{ 3 } \int { sin(u) } .du=\frac { 1 }{ 3 } (-cos(u))+C\\ -\frac { 1 }{ 3 } cos(x^{ 3 })+C
$