
Integral
calculus 

The fundamental theorem of
calculus 
The fundamental theorem of differential calculus 
Evaluating definite integrals
using indefinite integrals 
The fundamental theorem of integral calculus 
Evaluating the definite integral examples 






The fundamental theorem of calculus 
The
theorem that states the relationship between integration and
differentiation, that is, between areas and tangent
lines, is called the fundamental theorem
of calculus 

The fundamental theorem of differential calculus 
If
f (x) is continuous on
closed interval [a,
b] and F
(x)
is defined to be 

then,
F
is differentiable on (a,
b)
such that F' (x)
= f (x) for all x
in (a,
b).
This means that 


Evaluating definite integrals
using indefinite integrals 
To
evaluate the definite integral 

we should find one primitive
function F(x)
or antiderivative of 

the function f(x),
and since the indefinite integral 

is a primitive
function of f (x)
then, as two 

primitives of the same function can differ only by a
constant, we can write 

To
find the value of the constant C
that belongs to the lower limit a,
substitute x = a
to both sides of the 
above equality, and
since 

then C
= 
F(a),
so that 


Therefore,
the
definite and indefinite integrals are related by the fundamental theorem
of calculus. 
This
result shows that integration is inverse of differentiation. 

The fundamental theorem of integral calculus 
If
f(x)
is integrable on the interval [a,
b]
and F(x)
is an antiderivative of f
on (a,
b),
then 

The
right side of the above equation we usually write 

so
that, 



Thus,
to evaluate the definite integral we need to find an atiderivative F
of f,
then evaluate F(x)
at x = b
and 
at
x
= a, and calculate the
difference F(b)

F(a). 

Evaluating
the definite integral examples 
Example:
Find
the area under the line f(x)
= x
+ 1
over the interval [1,
5]. 
Solution:
To evaluate the
definite integral we need to find the atiderivative
F
of f
(x)
= x
+ 1, evaluate F(x) 
at x
= 5
and at x
= 1,
and calculate the difference 
F(5)

F(1). 





Cavalieri
 Gregory formula for quadrature of the parabola 
Example:
Let define the
surface area enclosed by the arc of the parabola f
(x)
= Ax^{2}
+ Bx
+ C
and xaxis
over the interval [a,
b]. 
Solution: 

where
the expression in the square brackets 




Thus,
the quadrature of the quadratic function leads to calculation of the
three ordinates or values of the function
f (a), 
f ((a+ b)/2)
and f
(b). 

Example:
Find
the area enclosed by the sine function f
(x)
= sin x
and the xaxis
over the interval [p/2,
3p/2].

Solution: 



Example:
Evaluate 



Solution: 




Example:
Find
the area enclosed by the graph of the cubic
f(x) = (1/3)x^{3}
+ 4x
and the positive part of the xaxis. 
Solution: The
limits of integrations are defined by the roots of the cubic that we can
find by solving the 
equation
f(x)
= 0, 




The
limits of integration, x_{1}
= 0 and x_{3}
= 2Ö3,
so 





Example:
Evaluate 



Solution: 












Calculus contents
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