Integration Definition
Opposite process to differentiation.
If ( f(t) ) is a function of ( t ), then ( g(t) = \frac{df}{dt} ) defines the derivative function.
For any constant ( C ):
[ g(t) = \frac{df(t)}{dt} = \frac{d}{dt}(f(t) + C) ]
Indefinite Integral
Defined as:
[ \int g(t) dt = f(t) + C ]
Thus,
[ \int \frac{df}{dt} dt = f(t) + C ]
Integral Notation
Symbol ( \int ) is a historical elongated ( S ) indicating a sum.
Example 1: Integration of 1
Find ( \int 1 dt ):
Here, ( g(t) = 1 ) gives ( f(t) = t ).
Result:
[ \int 1 dt = t + C ]
Example 2: Integration of t
Find ( \int t dt ):
Here, ( g(t) = t ) and ( f(t) = \frac{1}{2} t^2 ).
Result:
[ \int t dt = \frac{1}{2} t^2 + C ]
Example 3: Generalizing Integration
Find ( \int t^n dt ) for all ( n ):
Using ( g(t) = t^n ) leads to ( f(t) = \frac{1}{n+1} t^{n+1} ) (for ( n
eq -1 )).
Result:
[ \int t^n dt = \frac{1}{n + 1} t^{n + 1} + C ]
For ( n = -1 ):
[ \int t^{-1} dt = \int \frac{1}{t} dt ]
Differential Equation for Velocity
Equation:
[ \frac{dv}{dt} = a ]
Integrating both sides gives
[ \int dv = \int a dt ]
Result:
[ v = at + C ]
Fixed using initial velocity ( v(t=0) = u ):
Leads to [ v = u + at ]
Position Formula
Starting from:
( ds/dt = u + at )
Integrate:
[ s = \int (u + at) dt = ut + \frac{1}{2} a t^2 + C ]
Assuming ( s(0) = 0 ) gives ( C = 0 )
Thus:
[ s = ut + \frac{1}{2} a t^2 ]
Third Relationship Derivation
From ( v = u + at ):
Rearranging:
[ t = \frac{v - u}{a} ]
Substituting into ( s ):
Leads to [ s = \frac{u(v - u)}{a} + \frac{1}{2} a \left(\frac{v - u}{a}\right)^2 ]
Resulting in:
[ s = \frac{1}{2} a (v^2 - u^2) ]
Thus:
[ v^2 = u^2 + 2as ]
Core Formulas for 1-Dimensional Uniform Acceleration:
[ s = ut + \frac{1}{2} at^2 ]
[ v = u + at ]
[ v^2 = u^2 + 2as ]
Note:
These formulas are applicable only to 1-dimensional motion under uniform (constant) acceleration.