Integral calculus
The definite and indefinite integrals
Evaluating definite integrals using indefinite integrals
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 antiderivative 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 antiderivative 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 = 2Ö3, so
 Example:   Evaluate
 Solution:
Functions contents G
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