We can summarize the results from the previous web page to the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus
If a function  f is nonnegative and continuous on the closed interval  [a,b] then
∫ a b f ( x ) = F ( b ) - F ( a )
where  F is any function such that  F ' (x) = f (x) for all  x in  [a,b].

Three comments regarding the Fundamental Theorem of Calculus are in order.
First, this theorem describes a means for evaluating a definite integral, not a procedure for finding antiderivatives.
Second, when applying this theorem we find it helpful to use the formulation

∫ a b f ( x ) d x = F ( x ) ] a b = F ( b ) - F ( a )

For instance, we write

∫ 1 3 x 3 d x = x 4 4 ] = ( 3 ) 4 4 - ( 1 ) 4 4 = 81 4 - 1 4 = 20

Third, we observe that the constant of integration C can be dropped from the antiderivatives, because

∫ a b f ( x ) d x = [ F ( x ) + C ] a b =
= [ F ( b ) + C ] - [ F ( a ) + C ] =
= F ( b ) - F ( a ) + C - C =
= F(b) - F(a)

In our development of the Fundamental Theorem of Calculus, we assumed the function  f  to be nonnegative on the closed interval  [a,b]. With this assumption, we defined the definite integral as an area. Now, with the Fundamental Theorem, we can extend our definition of definite integrals to include functions that are negative on all part of the closed interval  [a,b].

Specifically, if  f  is any function that is continuous on the closed interval  [a,b], then the definite integral of  f(x) 
from a to b is given by
∫ a b f ( x ) = F ( b ) - F ( a )

where  F is an antiderivative of  f.

For instance,

∫ - 1 1 ( x 2 - 1 ) d x = x 3 3 - x ] - 1 1 =
= ( 1 3 - 1 ) - ( - 1 3 + 1 ) = - 4 3

Note that since the value of this particular definite integral is negative, it does not represent the area.

We now list some useful properties of the definite integral. In each case we assume that  f  and  g  are continuous on  [a,b].

Properties of Definite Integrals
1.    ∫ a b   k   f ( x )   d x   =   k   ∫ a b   f ( x )   d x ,   k  is a constant
2.    ∫ a b   [ f ( x )   ±   g ( x ) ]   d x   =   ∫ a b   f ( x )   d x     ±     ∫ a b   g ( x )   d x
3.    ∫ a b   f ( x )   d x   =   ∫ a c   f ( x )   d x   +   ∫ c b   g ( x )   d x ,   a < c < b
4.    ∫ a a   f ( x )   d x   =   0
5.    ∫ b a   f ( x )   d x   =   -   ∫ a b   f ( x )   d x


You can look for the more examples and test your ability for solving these problems.