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    Math Tutor USA - Algebra - Geometry

    Integration Made Easy

    By Masood Amir


    Calculus is a tool used to measure change or variation of a function with respect to the independent variable.

    Differential Calculus:

    It is the branch of calculus used to measure change or variation of a function in a very small interval of time, the techniques use to measure such changes is called “differentiation”.

    Integral Calculus:

    It is the branch of calculus used to measure changes or variation over an interval of independent variable, e.g to find length of curve, the area of region and the volume of a solid in a specified period of time.

    The technique used to measure such changes or variation is called “Integration” or “Antiderivatives”. It a reverse process of differentiation.

    Mathematically, Integration is defined as “ If f’(x) represents the differential coefficient of f(x), then the problem of integration is given f’(x), find f(x) or given dy/dx, find y.

    Notation: ”∫” is used to show the integration, it is a symbol of “S” derived from the word “Sum”. i.e. Integration is a process in which we have to sum up the derivatives over a specified interval and to find the function.

    Techniques of Integration:

    As we know that integration is the reverse process of differentiation, our problem is to find the function f(x) or Y, when f’(X) or dy/dx is given.

    dy/dx = f’(X)

    ∫dy = ∫f ’(X)dx

    Y= f(x) is our solution

    Ist Formula of Integration (1st Rule of Integration)

    Indefinite Integration:

    Ist Formula of Integration (Ist Rule of Integration):

    Let ∫ dy = ∫ xndx

    y = xn+1/n+1  + C

    Why “C”:

    In the process of differentiation, we eliminate constant, as the derivative of a constant is “zero”.

    So, In functions like Xn, Xn+ 6, Xn + 3 ,  Xn – K, the derivatives of all of them is Xn-1, in finding the anti derivative of  Xn-1we put a constant “C”, as we don’t know which constant was present in the original function, and this can be found If we have initial boundary values (Definite Integral).

    Example: Solve ∫x3dx

    Solution:        x3+1/3+1 + C   = x4/4 + C

    Example: Solve ∫(x+ x2 + 5x + 6)dx

    Solution:      ∫x3dx + ∫x2dx + ∫5xdx + ∫6dx

                             X4/4 + x3/3 + 5x2/2+ 6x + C

    Worksheet # 1


    Find the Integral of the following:

                     SET 1                                         

    ∫(x3-4x2+5x-6)dx                        ∫(3x5-4x3+3x2)dx

    ∫(ax5-bx4)dx                               ∫(x3/2-5x4/3+3x2)dx

    ∫(4√3x2 -2x)dx                            ∫(3x + 5x2 –x3/2-0.4x4)dx

    ∫(x(8x-1/2)dx                               ∫(2-x)(4+3x)dx

    ∫(x-3 + x-4)dx                               ∫(2x3-3)3x4dx

    ∫(4x7 + 3x12 -5x8 + 2x -1)dx          ∫(ax3 – bx2 + cx –d)dx

    ∫(1/x3 + 2/x2 -6)dx                        ∫(-3x-8 + 2√x)dx

    ∫((x3 -5)(2x + 5)dx                         ∫(7x-6 + 5√x)dx