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    عدد المساهمات : 116
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    تاريخ التسجيل : 08/03/2013
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    مُساهمةموضوع: Integration   الخميس مارس 21, 2013 6:07 pm


    Integration
    Integration is used to find areas under curves.
    Integration is the reversal of differentiation hence functions can be integrated
    by indentifying the anti-derivative.
    However, we will learn the process of integration as a set of rules rather than
    identifying anti-derivatives.
    Terminology
     Indefinite and Definite integrals
    There are two types of integrals: Indefinite and Definite.
    Indefinite integrals are those with no limits and definite integrals have
    limits.
    When dealing with indefinite integrals you need to add a constant of
    integration. For example, if integrating the function f(x) with respect to x:
    ( )

    f x dx = g(x) + C
    where g(x) is the integrated function.
     C is an arbitrary constant called the constant of integration.
     dx indicates the variable with respect to which we are integrating, in this
    case, x.
     The function being integrated, f(x), is called the integrand.Integration- the basics
    2
    The rules
     The Power Rule

    x dx
    n
    = C
    n 1
    x
    n 1
    +
    +
    +
    provided that n ≠ -1
    Examples:

    x dx
    5
    =
    6
    x
    6
    + C


    x dx
    -4
    =
    - 3
    x
    -3
    + C
     When n = -1

    x dx
    -1
    =

    dx
    x
    1
    = ln x + C
     Constant rule

    k dx = kx + C where k is a constant
    Example:

    2dx = 2x + C
     Exponentials

    e dx
    kx
    = e C
    k
    1 kx
    +
    Example: e C
    9
    1
    e dx
    9x 9x
    = +

    e dx e C
    x x
    = +

     Trig functions
    - Cos
    ( ) =

    cos x dx sin(x) + C
    ( ) =

    cos kx dx sin( ) kx
    k
    1
    + C where k is a constant
    Example: ( ) =

    cos 12x dx sin( ) 12x
    12
    1
    + C Integration- the basics
    3
    - Sin
    ( ) =

    sin x dx -cos(x) + C
    ( ) =

    sin kx dx cos( ) kx
    k
    1
    − + C where k is a constant
    Example: ( ) =

    sin 10x dx cos( ) 10x
    10
    1
    − + C
    ( ) =

    sin - 5x dx cos( ) - 5x
    5
    1
    + C
    Linearity
    Suppose f(x) and g(x) are two functions in terms of x, then:
    [ ( ) ( )] ( ) ( )
    ∫ ∫ ∫
    f x ± g x dx = f x dx ± g x dx
    Additionally, if A and B are constants, then
    [ ( ) ( )] ( ) ( )
    ∫ ∫ ∫
    Af x ±Bg x dx = A f x dx ± B g x dx
    Examples:
    ( )

    2x + 3x dx
    4 5
    =

    2x dx
    4
    +

    3x dx
    5
    =

    2 x dx
    4
    +

    3 x dx
    5
    =




    5
    x
    2
    5
    +




    6
    x
    3
    6
    + C
    =
    2
    x
    5
    2x
    5 6
    + + C

    ( ( ) )

    5cos 3x − 3e dx
    7x
    = ( )
    ∫ ∫
    5cos 3x dx - 3e dx
    7x
    = ( )
    ∫ ∫
    5 cos 3x dx - 3 e dx
    7x
    = ( ) 





     −




     7x
    e
    7
    1
    sin 3x 3
    3
    1
    5
    = ( )
    7x
    e
    7
    3
    sin 3x
    3
    5
    −Integration- the basics
    4
    Questions (General rules):
    Integrate the following functions:
    1. ( )

    x − x + 5 dx
    2
    3
    x
    6 1
    2. ( )

    3x + x − 5 dx
    8
    3. ( )

    9x − 3x dx
    2 -1
    4. ( ( ) )

    sin 4x + e dx
    3x
    5. ( ( ) )

    cos 7x + 7x dx
    2
    (Solutions on page Cool
    Definite Integrals
    Earlier we saw that
    ( )

    f x dx = g(x) + C
    Suppose now we are given limits, i.e.
    ( )

    b
    a
    f x dx = g(x) + C
    This can be interpreted as:
    (value of g(x) + C at x = b) – (value of g(x) + C at x = a)
    In other words, since C will cancel out:
    ( )

    b
    a
    f x dx = g(b) – g(a)
    The full calculation of definite integrals is usually written out as:
    ( )

    b
    a
    f x dx = [ ( )]
    b
    a g x = g(b) – g(a)
    i.e. integrate the function first (find g(x)) then substitute in the given limits
    (always substitute the upper limit first).
    (where a is the lower limit
    and b is the upper limit) Integration- the basics
    5
    Examples
    1.

    1
    0
    2
    x dx =
    1
    0
    3
    x
    3
    1






    = [ ]
    1
    0
    3
    x
    3
    1
    =
    3
    1
    {(1)
    3
    – (0)
    3
    } =
    3
    1
    (1 – 0) =
    3
    1
    2. ( ) 2x 1 dx
    3
    ∫1
    + =
    3
    1
    2
    x
    2
    2x






    + = [ ]
    3
    1
    2
    x + x = {(3
    2
    + 3) – (1
    2
    + 1)}
    = {(9 + 3) – (1 + 1)} = 12 – 2 = 10
    3. cos ( ) x dx
    2
    π
    ∫0
    = [ ( )]
    2
    π
    0
    sin x = {(sin(
    2
    π
    )) – (sin(0))}
    = 1 – 0 = 1
    Questions (Definite integrals):
    Integrate the following functions:
    1. (3x 2x 5)dx
    2
    1
    2

    − +
    2. e dx
    1
    0
    7x

    3. ( )

    π
    0
    sin 2x dx
    4. ( )

    +
    4
    1
    4x
    12e 4 x dx
    (Solutions on page 9)
    Integration that leads to log functions
    We know that if we differentiate y = ln(x) we find
    x
    1
    dx
    dy
    = .
    We also know that if y = ln f(x), this differentiates as:
    ( )
    f( ) x
    'f x
    dx
    dy
    =
    If we can recognise that the function we are trying to integrate is the derivative
    of another function, we can simply reverse the above process. So if the
    function we are trying to integrate is a quotient, and if the numerator is the
    derivative of the denominator, then the integral will involve a logarithm, i.e.
    ( )
    ( )

    dx
    f x
    'f x
    = ln (f(x)) + C
    Example: y = ln(2x
    2
    + 5)
    t = 2x
    2
    + 5 y = ln t
    dx
    dt
    = 4x
    dt
    dy
    =
    t
    1
    dx
    dy
    = 4x x
    t
    1
    =
    t
    4x
    =
    + 5
    2
    2x
    4xIntegration- the basics
    6
    Example 1:

    +
    dx
    3 5x
    5
    The derivative of the denominator is 5 which is the same as the
    numerator, hence

    +
    dx
    3 5x
    5
    = ln (3 + 5x) + C
    Example 2:

    +
    dx
    1 x
    x
    2
    The derivative of the denominator is 2x. This is not the same as the
    numerator but we can make it the same by re-writing the function
    2
    1 x
    x
    +
    as
    2
    1 x
    2x
    2
    1
    +
    ⋅ , therefore

    +
    dx
    1 x
    x
    2
    =

    +
    dx
    1 x
    2x
    2
    1
    2
    =
    2
    1
    ln (1 + x
    2
    ) + C
    Example 3:
    ( )

    dx
    xln x
    1
    The derivative of ln x is
    x
    1
    , so we can rewrite the function as:
    ln( ) x
    x
    1
    . Hence
    ( )

    dx
    xln x
    1
    =
    ( )

    dx
    ln x
    x
    1
    = ln(ln(x)) + C
    Example 4:







    +

    2
    1
    dx
    x 1
    3
    x
    3







    +

    2
    1
    dx
    x 1
    3
    x
    3
    =







    +
    2
    1
    dx
    x 1
    1
    -
    x
    1
    3
    = [ ( ) ( )]
    2
    3ln x − 3ln x +1 1
    = {(3ln(2) – 3ln(3)) – (3ln(1) – 3ln(2))}
    = 3ln(2) – 3ln(3) + 3ln(2) = 6ln(2) – 3ln(3)
    = ln(2
    6
    ) – ln(3
    3
    )
    = ln(64) – ln(27) = 





    27
    64
    lnIntegration- the basics
    7
    Questions (Integration that leads to log functions):
    Integrate the following functions:
    1.

    +
    dx
    2 3x
    3
    2.

    +
    dx
    1 2x
    x
    2
    3.

    +
    dx
    e 1
    e
    2x
    2x
    4.

    +
    dx
    x 4
    x
    2-
    -3
    5.







    +

    +
    1
    0
    dx
    x 2
    1
    x 1
    1
    (Solutions on page 10) Integration- the basics
    8
    Solutions (General rules):
    1. ( )

    x − x + 5 dx
    2
    3
    x
    6 1
    =
    ∫ ∫ ∫
    x dx - x dx + 5 dx
    2
    3
    x
    6 1
    =
    ∫ ∫ ∫

    x dx - x dx + x dx
    6 2 5
    3
    =
    ( ) 4
    x x
    7
    x
    4
    2
    5
    7 2
    5

    − +

    + C
    =
    4
    7
    4x
    1
    5
    2x
    7
    x
    2
    5
    − − + C
    2. ( )

    3x + x − 5 dx
    8
    =

    3x dx
    8
    +

    x dx -

    5dx
    =

    3 x dx
    8
    +

    x dx -

    5dx
    = 5x
    2
    x
    9
    3x
    9 2
    + − + C
    = 5x
    2
    x
    3
    x
    9 2
    + − + C
    3. ( )

    9x − 3x dx
    2 -1
    =

    9x dx
    2
    -

    3x dx
    -1
    =

    9 x dx
    2
    -

    3 x dx
    -1
    =
    3
    9x
    3
    - 3ln + C (x)
    =
    3
    3x - 3ln + C (x)
    4. ( ( ) )

    sin 4x + e dx
    3x
    = ( )

    sin 4x dx +

    e dx
    3x
    = ( )
    3x
    e
    3
    1
    cos 4x
    4
    1
    − + + C
    5. ( ( ) )

    cos 7x + 7x dx
    2
    = ( )

    cos 7x dx +

    7 x dx
    2
    = ( )
    3
    x
    3
    7
    sin 7x
    7
    1
    + + C Integration- the basics
    9
    Solutions (Definite integrals):
    1. (3x 2x 5)dx
    2
    1
    2

    − + =
    2
    1
    3 2
    5x
    2
    2x
    3
    3x






    − +
    = [ ]
    2
    1
    3 2
    x − x + 5x
    = {(2
    3
    – 2
    2
    + 5(2)) – (1
    3
    – 1
    2
    + 5(1))}
    = {(8 – 4 + 10) – (1 – 1 + 5)}
    = 14 – 5
    = 9
    2. [ ]
    1
    0
    7x
    1
    0
    7x
    1
    0
    7x
    e
    7
    1
    e
    7
    1
    e dx =






    =

    =
    7
    1
    {e
    7
    – e
    0
    } =
    7
    1
    (e
    7
    – 1)
    3. ( )

    π
    0
    sin 2x dx = ( )
    π
    0
    cos 2x
    2
    1






    − = [ ( )]
    π
    2 0
    1
    − cos 2x
    = -
    2
    1
    {cos(2π) – cos (0)}
    = -
    2
    1
    {1 – 1}
    = 0
    4. ( )

    +
    4
    1
    4x
    12e 4 x dx = ( )

    +
    4
    1
    4x
    12e 4x dx
    2
    1
    =
    4
    1
    2
    3
    4x 2
    3
    4x
    4
    12e




    + =
    4
    1
    4x
    3
    8x
    3e
    2
    3




    +
    =
    ( )
















    − +




    +
    3
    8
    3e
    3
    8 4
    3e
    16 4
    2
    3
    =
    ( )












     − +





    +
    3
    8
    3e
    3
    8 8
    3e
    16 4
    =
    3
    8
    3e
    3
    64
    3e
    16 4
    + − −
    =
    3
    56
    3e 3e
    16 4
    − +Integration- the basics
    10
    Solutions (Integration that leads to log functions):
    1.

    +
    dx
    2 3x
    3
    = ln (2 + 3x) + C
    2.

    +
    dx
    1 2x
    x
    2
    Differentiating the denominator gives 4x
    Therefore rewrite the function:
    2
    1 2x
    x
    +
    =
    2
    1 2x
    4x
    4
    1
    +

    Hence,

    +
    dx
    1 2x
    x
    2
    =

    +
    ⋅ dx
    1 2x
    4x
    4
    1
    2
    =

    +
    dx
    1 2x
    4x
    4 2
    1
    =
    4
    1
    ln (1 + 2x
    2
    ) + C
    3.

    +
    dx
    e 1
    e
    2x
    2x
    Differentiating the denominator gives 2e
    2x
    hence we can rewrite the
    function as:
    e 1
    e
    2x
    2x
    +
    =
    e 1
    2e
    2
    1
    2x
    2x
    +


    +
    ⋅ dx
    e 1
    e
    2
    1
    2x
    2x
    =
    2
    1
    ln (e
    2x
    + 1) + C
    4.

    +
    dx
    x 4
    x
    2-
    -3
    Differentiating the denominator gives -2x
    -3
    , hence the function can be
    rewritten as:
    x 4
    x
    2-
    -3
    +
    =
    x 4
    2x
    2
    1
    2-
    -3
    +
    − ⋅

    +
    dx
    x 4
    x
    2-
    -3
    =

    +
    − dx
    x 4
    2x
    2
    1
    2-
    -3
    = -
    2
    1
    ln(x
    -2
    + 4) + C
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