Integral calculus
      The definite and indefinite integrals
      The area between the graph of a function and the x-axis over a closed interval
         Geometric interpretation of the definite integral
      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 area between the graph of a function and the x-axis over a closed interval
Geometric interpretation of the definite integral
Until now we assume that the integrand, the function that is integrated, to be nonnegative or  f (x) > 0 for all x in an interval [a, b].
Now suppose f (x) < 0 in the whole interval or in some of its parts then, the areas of regions between the graph of  f and the x-axis, which lie below or above the x-axis, differ in the sign of  f (x).
 
Therefore, the definite integral yields the algebraic sum of these areas taking regions below the x-axis negative, as show the figures above.
So, if the graph of  f looks as in the left figure above then
hence the definite integral represents the algebraic sum of the areas above and below the x-axis.
Thus, as the right figure above shows, follows that
as the area A2, lying under the arc of the sinusoid in the interval [p, 2p] is congruent to the area A1 in [0, p] but with opposite sign.
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) = Ax2 + Bx + C and x-axis 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 x-axis 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)x3 + 4x and the positive part of the x-axis.
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, x1 = 0 and x3 = 23, so
 
Example:   Evaluate  
Solution:     
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