Integration




Differentials

Differentiation

With the derivative f'(x) of function f(x), we construct a new function.
df(x) = f'(x).h
Here h is an arbitrary non zero number .
This new function is called the differential function of f(x).

Trivial examples

The differential function of f(x)= (6x.x + 7) is d(6x.x + 7) = 12x.h
The differential function of f(x)= sin(x) is d sin(x) = cos(x).h
The differential function of f(x)= x is dx = h

New notation

Because of that last example, we see that there is no difference between dx and h.
Hence, we can write the definition of df(x) as
 
        df(x) = f'(x).dx

Formula df(u)

Say u = g(x), then du = g'(x).dx = u'. dx
Now, take the function f(u) as a function of x. Then we have
 
        df(u) = f'(u).u'.dx = f'(u).du
example:

        d sin(sqrt(x)) = cos(sqrt(x)).d sqrt(x)
 
df(x) = f'(x).dx   and  df(u) = f'(u).u'.dx = f'(u).du

Indefinite integral

Root functions or primitive functions

A function F(x) is called a root function or primitive function of f(x) if and only if F'(x) = f(x).
Example:
-cos(x) is a primitive function of sin(x). -cos(x) + 12 is also a primitive function of sin(x).

So, if F(x) is primitive function of f(x), then F(x) + an arbitrary constant is also a primitive function of f(x).
All these primitive functions of f(x) are written as F(x) + C
F(x) is a root function or primitive function of f(x)

if and only if

F'(x) = f(x).


Definition of indefinite integral

Say F(x) is an arbitrary primitive function of f(x), then dF(x) = f(x)dx.
So, all primitive functions of f(x) have the same differential.
Therefore, we say : ' the indefinite integral of f(x)dx is F(x) +C '.
The constant number C reminds us of all the primitive functions of f(x).
We write
 
        /
        | f(x)dx = F(x) + C
        /
The integral sign is clumsy due to the html file, but that is not essential.
Example:
 
        /
        | sin(x)dx = -cos(x) + C because d(-cos(x)) = sin(x)dx
        /
 
        /
        | f(x)dx = F(x) + C  <=> d F(x) = f(x)dx
        /

Formulas

In the formulas are u and v functions of x.
 

        /  r       ur+1
        | u du  = ------- + C     (for all real r except r = -1)
        /          r+1


 
Proof:
            r+1
           u        1
        d ------ = ----. (r+1).ur  .u'.dx = ur .du
           r+1     r+1
In a similar way you can prove the following formulas.
 
        /   1             ___
        | -------du  = 2.V u  + C
        |    ___
        /   V u

        /
        | eu du = eu  + C
        /

        /  u       au
        | a  du = ----- + C
        /         ln(a)

        /
        | sin(u)du = -cos(u) + C
        /

        /
        | cos(u)du = sin(u) + C
        /

        /     1
        | ---------du = -cot(u) + C
        | sin2 u
        /


        /     1
        | ---------du =  tan(u) + C
        |  cos2 u
        /


        /     1
        | ---------du =  arctan(u) + C
        |    1+u2
        /


        /     1
        | --------------du =  arcsin(u) + C
        |    ________
        |   |      2
        /  \| 1 - u


        /  1
        | ---du = ln|u| + C
        /  u


        /            /
        | a.u.dx = a.| u.dx   (with a constant)
        /            /


        /           /       /
        | (u+v)dx = | udx + | vdx
        /           /       /


Examples

 
        /
        | x6 x = x7 /7 + C
        /

        /  dx
        | ------ = x-2/(-2) + C
        /  x3

        /
        | (x+3)dx = x2 /2  + 3x + C
        /

        /
        | 3x dx = 3x /ln(3)  + C
        /

        /      7      /
        | (x+2) dx =  | (x+2)7 d(x+2) = (x+2)8 /8 + C   (since d(x+2)=dx)
        /             /

        in the same way

        /  x+2     /
        | 3    dx= | 3x+2 d(x+2) = 3x+2/ln(3)  + C
        /          /

        /
        | 5.cos(x)dx = 5 sin(x) + C
        /

        Since d(5x) = 5.dx

        /             1  /                1
        | sin(5x)dx = -. | sin(5x)d(5x) = -.(-cos(5x)) + C
        /             5  /                5

        Since d(5x+3) = 5.dx

        /       7      1  /        7           1        8
        | (5x+3) dx =  -. |  (5x+3) d(5x+3) = --. (5x+3)  +C
        /              5  /                   40

Integration using substitution

If there is a part g(x) in the integral such that g'(x).dx is in that integral too, then do the substitution g(x) = u.

Examples :

Another 3 formulas


 
Proof :

                        2          1                       2u
    d ln(|u + sqrt(k + u ) |) = -----------------( 1 + ---------------)u'dx
                                              2                     2
                                u + sqrt(k + u )        2.sqrt(k + u )


             1                 u + sqrt(k + u2)
        = -----------------( -------------------) du
          u + sqrt(k + u2)     sqrt(k + u2)


               1
        =  ------------- du
              ________
             |      2
            \| k + u
Example:
 
        /     dx          /     dx            /     d(x + 1 )
        | ------------- = | --------------- = | -------------
        /  x.x + 2x + 3   |         2         |         2
                          / (x + 1 )  + 2     / (x + 1 )  + 2

            1            x + 1
         = ---- * arctan(-----) + C
            ___           ___
           V 2           V 2


Integration by parts

Formula :
 
        /              /
        | u dv = u.v - | v du
        /              /

 
Proof:

              /
    d ( u.v - | v du ) = ... = u dv
              /
Examples:

Integration of a rational fraction

Partial fraction decomposition

If N(x) and D(x) are polynomials, then N(x)/D(x) is called a rational fraction.

Each rational fraction can be written as the sum of a polynomial and partial fractions. This is called the partial fraction decomposition. Partial fraction decomposition is essential to integrate rational fractions.

The procedure is explained here as a separate subject.

Integration of partial fractions

If we can integrate each partial fraction, we can integrate all rational fractions.
Now we'll show how to integrate each partial fraction.
Each rational fraction can be written as the sum of a polynomial and a sum of partial fractions.
Each polynomial and each partial fraction can be integrated.
Therefore, each rational fraction can be integrated in the way as shown above.

Integration of trigonometric functions

product of sin and cos

Integration of the forms :
 
/                       /                      /
| cos(rx).cos(sx) dx ;  | sin(rx).sin(sx) dx ; | sin(rx).cos(sx) dx
/                       /                      /

We appeal on the trigonometric formulas, deduced from the Simpson formulas
 
        cos((r + s)x) + cos((r - s)x) = 2cos(rx)cos(sx)
        cos((r + s)x) - cos((r - s)x = - 2sin(rx)sin(sx)
        sin((r + s)x) + cos((r - s)x) = 2sin(rx)cos(sx)
Example :
 
/                             /
| sin(x).sin(7x) dx = (-1/2). | (cos(8x) - cos(-6x))dx
/                             /


           /                     /
=  (-1/2). | (cos(8x) dx + (1/2).| (cos(6x) dx
           /                     /

=  (-1/16).sin(8x) +(1/12).sin(6x) + C

product of powers of sin and cos

Integration of the form :
 
/
| sinm (u).cosn (u) dx
/

Here n and m are positive integers.
If n is odd, then let sin(u) = t
If m is odd, then let cos(u) = t
If m and n are even, then use the Carnot formulas and/or the formula sin(2x) = 2.sin(x)cos(x).
Examples :

Rational functions of sin(u) and cos(u)

To integrate a rational function of sin(u) and cos(u) use the t-formulas.

 
Let t = tan(u/2) , then

                2
           1 - t                  2t                   2t
cos(u) = ---------  ; sin(u) =  -------- ; tan(u) =  -------
                2                     2                    2
           1 + t                 1 + t                1 - t

and                                   dt
        u/2 = arctan(t)  => du = 2 ---------
                                         2
                                    1 + t
Examples :

Integration of irrational functions

With a suitable substitution

Sometimes the integration of an irrational function is possible with the help of a suitable substitution.

Examples :

Special sort of irrational functions

 
Now we'll show how to integrate the forms

      /      _____________
      |     |  2
      |    \| x  + b x + c  dx
      /
and

      /    _____________
      |   |            2
      |  \| c + b x - x  dx
      /

 
First consider

     /   ________
     |  |  2
I =  | \| t  + k  dt   with k real and constant.
     /

To integrate this we use integration by parts
            ________
           |  2
Let u =   \| t  + k   and  dt = dv  then


                    /
      ________      |
     |  2           |          t
I = \| t  + k  .t - |t. ---------------dt
                    |       ________
                    |      |  2
                    /     \| t  + k


                    /
      ________      |   2
     |  2           |  t  + k - k
I = \| t  + k  .t - |---------------dt
                    |       ________
                    |      |  2
                    /     \| t  + k

                    /                      /
      ________      |   2                  |
     |  2           |  t  + k              |      1
I = \| t  + k  .t - |-------------- dt + k |---------------- dt
                    |       ________       |       ________
                    |      |  2            |      |  2
                    /     \| t  + k        /     \| t  + k

                    /                      /
      ________      |   ________           |
     |  2           |  |  2                |      1
I = \| t  + k  .t - | \| t  + k  dt    + k |---------------- dt
                    |                      |       ________
                    |                      |      |  2
                    /                      /     \| t  + k


                          /
        ________          |
       |  2               |      1
2.I = \| t  + k  .t  +  k |---------------- dt
                          |       ________
                          |      |  2
                          /     \| t  + k

        ________                  ________
       |  2                      |  2
2.I = \| t  + k  .t + k ln |t + \| t  + k |  + C

        ________                  ________
       |  2                      |  2
      \| t  + k  .t + k ln |t + \| t  + k |  + C
 I = --------------------------------------------
                        2


With this formula we can integrate

      /      _____________
      |     |  2
      |    \| x  + b x + c  dx
      /

Example :
        /  _____________       /   _____________
        | |  2                 |  |        2
  I =   |\| x  + 4 x + 6 dx  = | \| (x + 2)  + 2  dx
        |                      |
        /                      /
Now let x+2 = t , then we have

          /   ________
          |  |  2
    I =   | \| t  + 2  dt
          /
        ________                  ________
       |  2                      |  2
      \| t  + 2  .t + 2 ln |t + \| t  + 2 |  + C
 I = --------------------------------------------
                        2

In a similar way we can integrate

     /   ________
     |  |      2
I =  | \| k - t   dt with k real and constant.
     /

and so all integrals of the form

      /    _____________
      |   |            2
      |  \| c + b x - x  dx
      /

With a trigonometric substitution

If the integrand contains
sqrt(a2 - u2 ) or sqrt(a2 + u2 ) or sqrt(u2 - a2 )
you can use a trigonometric substitution.
 
For sqrt(a2 - u2 )  use the substitution   u = a.sin(t)
    with t in [-pi/2,pi/2]


For sqrt(a2 + u2 )  use the substitution   u = a.tan(t)
    with t in [-pi/2,pi/2]


For sqrt(u2 - a2 )  use the substitution   u = a.sec(t)
        with t in [0, pi/2[ if u > 0  and
        with t in [ pi, 3.pi/2] if u < 0

We give for each case an example.

Online Integrator

Via this link you'll find a powerful online integrator.

Definite integral

Definition

Say f(x) is continuous in [a,b]. In that interval we choose values x1,x2,x3,...,xn-1. Take a = x0 and b = xn.
So, we have n partial intervals in [a,b]. The value (xi - xi-1) is the width of the i-th interval.
In the i-th interval, f(x) reaches a maximum value Mi and a minimum value mi.
We create the 'lower sum'
 
        s = sum mi.(xi - xi-1)
             i
and the 'upper sum'
 
        S = sum Mi.(xi - xi-1)
             i
Now we choose in each interval [xi-1, xi] in new value. In that way, we create 2n intervals in [a,b]. With these intervals corresponds a new lower sum and a new upper sum.
In that way we proceed incessant with doubling the number of intervals.
Then, we have a sequence of lower sums, and a sequence of upper sums.
The sequence of the lower sums is not descending.
The sequence of the upper sums is not rising.
Besides each upper sum is greater or equal to each lower sum.
Hence, the sequence of the lower sums is a not descending and upper bounded sequence has a finite limit. The sequence of the upper sums is a not rising and lower bounded sequence has a finite limit. It can be proved that these limits are the same.
This limit I is called the definite integral of f(x)dx.
The number b is called the upper limit, and a is called the lower limit of the definite integral. All this is written as
 
        /b
        |  f(x).dx = I
        /a

Numeric example

Take the function y = x2 in [0,1]. We divide the interval in n equal parts.
 
        n               lower sum               upper sum
        10              0.317                   0.385
        20              0.325                   0.358
        40              0.329                   0.346
        80              0.331                   0.340
        160             0.332                   0.336
        ...
        2560            0.33327                 0.3335
Both sums converge to the same limit I = 1/3.
 
        /1
        |  x2dx  =  1/3
        /0

Properties

Definite integral as Riemann sum

We choose in each interval [xi-1,xi] an x-value = si. Then
 
    xi-1 =<    si    =< xi

=>     mi  =<   f(si)  =< Mi

=>    mi.(xi - xi-1) =< f(si).(xi - xi-1) =< Mi.(xi - xi-1)

We make the sum of these expressions for i from 1 to n.
 
sum mi.(xi - xi-1) =< sum f(si).(xi - xi-1) =< sum Mi.(xi - xi-1)
The sum f(si).(xi - xi-1) is called a Riemann sum.

We take the limit for n --> infinity

 
           I =< lim sum f(si).(xi - xi-1) =< I
Therefore
 
        /b
    I = |  f(x).dx  =  lim sum f(si).(xi - xi-1)
        /a
Each definite integral is the limit of an appropriate Riemann sum and each limit of an appropriate Riemann sum is a definite integral.

Theorem of the mean.

If f(x) is continuous in [a,b] then, there is at least one c in [a,b] such that
 
        /b
        |  f(x).dx = (b-a).f(c)
        /a
Proof :
Say m is the smallest and M the biggest value of f(x) in [a,b].
In [a,b] we choose values x1,x2,x3,...,xn-1. As in the definition, again, we make lower and upper sums. In all sums i goes from 1 to n.
 
        sum m.(xi - xi-1) =< sum mi.(xi - xi-1)


<=>     m.sum (xi - xi-1) =< sum mi.(xi - xi-1)


and
        sum Mi.(xi - xi-1) >= sum M.(xi - xi-1)


        sum Mi.(xi - xi-1) >= M.sum (xi - xi-1)

Taking the limit, we have
 
        m.(b-a) =< I =< M.(b-a)
Hence I/(b-a) is a number between the smallest and the biggest image in [a,b].
But, since f(x) is continuous in [a,b], there is a value c in [a,b], such that f(c) = I/(b-a) .
Hence, there is at least one c in [a,b] such that
 
        /b
        |  f(x).dx = (b-a).f(c)
        /a

Mean value of a function in [a,b]

The value f(c) from the previous theorem is called the mean value of f(x) in [a,b].
 
                                            1     /b
        The mean value of f(x) in [a,b] = ------- |  f(x).dx
                                          (b - a) /a

The name x in a definite integral.

 
        /b
        |  f(x).dx = (b-a).f(c) = number independent of x
        /a
So, the name x is not important. Hence,
 
        /b           /b           /b
        |  f(x).dx = |  f(u).du = |  f(t).dt = ...
        /a           /a           /a

Definite integral with variable upper limit

f(x) is continuous in [a,b] and x is a value in [a,b]. The integral
 
        /x
        |  f(t).dt
        /a
is a number depending on the upper limit x. It is a function of x. We call that function g(x). Hence
 
        /x
 g(x) = |  f(t).dt
        /a

First fundamental theorem

The derivative of the function
 
          /x
          |  f(t).dt
          /a
is f(x) .

 
Let
        /x
 g(x) = |  f(t).dt
        /a
We calculate the derivative of g(x) appealing on the definition of derivative.
 
        d               (g(x+h)-g(x))
        -- g(x) =  lim ---------------
        dx        h->0        h

                        /x+h            /x
                        |   f(t).dt  -  |  f(t).dt
                        /a              /a
                =  lim -----------------------------
                  h->0                h


                        /x+h            /a
                        |   f(t).dt  +  |  f(t).dt
                        /a              /x
                =  lim -----------------------------
                  h->0                h


                        /x+h
                        |   f(t).dt
                        /x
                =  lim -------------
                  h->0      h

        The theorem of the mean says that there is at least one c
        in [x,x+h] such that

                        h.f(c)
                =  lim -------------
                  h->0      h


                =  lim f(c)
                  h->0

        If h -> 0 , c -> x

                =  lim f(c)
                  c->x

                = f(x)

Corollary

Since
 
   d    /x                      /x
   ---  |  f(t).dt = f(x)  ,    |  f(t).dt is a primitive function of f(x)
   dx   /a                      /a

Second fundamental theorem

If F(x) is a primitive function of f(x), then
 
        /b
        |  f(t).dt  = F(b) - F(a)
        /a

Proof: F(x) is a primitive function of f(x) and
 
        /x
        |  f(t).dt is a primitive function of f(x) too.
        /a
Thus,
        /x
        |  f(t).dt = F(x)  + C    FOR ALL x in [a,b]    (1)
        /a

For x = a this gives:

        /a
   0 =  |  f(t).dt = F(a)  + C    FOR ALL x in [a,b]
        /a

=>      C = - F(a)

        /x
(1) =>  |  f(t).dt = F(x)  - F(a)   FOR ALL x in [a,b]
        /a

For x = a this gives:

        /b
        |  f(t).dt = F(b)  - F(a)
        /a

Since the name t is not important,

        /b
        |  f(x).dx = F(b)  - F(a)
        /a

The last expression is noted as

        /b                    b
        |  f(x).dx =  [ F(x) ]
        /a                    a

Calculation of a definite integral

Example:
 
        /1
        |  x(x2  + 7)dx
        /0

We calculate a primitive function of x(x2  + 7).

        /
        |  x(x2  + 7)dx  = ... = (1/4)(x2  + 7)2  + C
        /
Now,
        /1    2                    2     2  1
        |  x(x  + 7)dx  = [ (1/4)(x  + 7)  ]  = 64/4 - 49/4 = 15/4
        /0                                  0

Application of definite integrals

Area between two curves.

The area enclosed between two continuous curves y = f(x) and y = g(x) in the interval [a,b] with f(x) >= g(x) in [a,b] is
 
   /b
   |  (f(x) - g(x))dx
   /a

Proof:

In interval [a,b], we choose values x1,x2,x3,...,xn-1. Take a = x0 and b = xn.
We choose in each interval [xi-1,xi] a value = si.
Consider the area of the rectangle in [xi-1,xi]

 
    ( f(si) - g(si) ).(xi - xi-1)
Now, take the sum of the area of such rectangles in each subinterval.
 
    sum ( f(si) - g(si) ).(xi - xi-1)
This is a Riemann sum. The limit a this sum , for n --> infinity, is the definite integral
 
   /b
   |  (f(x) - g(x))dx
   /a
This is the area enclosed between the continuous curves y = f(x) and y = g(x).

You see that the area is the sum of an infinity number of elementary rectangles. This procedure is used to solve a lot of problems in physics.

If the condition f(x) >= g(x) is not satisfied in [a,b], you can divide the interval in a suitable way such that the condition is satisfied in each subinterval.

Example:
We calculate the area between y = cos(x) and y = sin(x) in [0,pi].

In [0,pi/4], we have cos(x) >= sin(x) and in [pi/4,pi] we have sin(x) >= cos(x).

The area A =

 
    /pi/4                        /pi
    |     (cos(x) - sin(x)) dx + |    (sin(x) - cos(x)) dx
    /0                           /pi/4

                       |pi/4                      |pi
    = (sin(x) + cos(x))|      + (-cos(x) - sin(x))|
                       |0                         |pi/4



    = ... = 2.8284

Area calculations

You can find area calculations through this link

Volume of bodies of revolution

A continuous function f(x) > 0 in interval [a,b], rotating around the x-axis, defines a volume.

We'll find a formula to calculate this volume V.

In interval [a,b], we choose values x1,x2,x3,...,xn-1. Take a = x0 and b = xn.
We choose in each interval [xi-1,xi] a value = si.
Consider the rectangle with base (xi - xi-1) and height si.
If this rectangle revolves around the x axis, the volume of the cylinder is

 
   pi. si2 . (xi - xi-1)
The sum of all such cylinders is a Riemann sum
 
    sum (pi. si2 . (xi - xi-1))
The limit a this sum, for n --> infinity, is the definite integral
 
      /b
  pi. |  (f(x))2 dx
      /a
A continuous function f(x) with positive images in interval [a,b], revolving around the x-axis, defines a volume
 
              /b
    V =   pi. |  (f(x))2 dx
              /a

Example:
When the curve of y = sqrt(r2-x2) revolves around the x-axis, it generates a sphere with radius r. The volume =

 
            /r
   V = pi . |   (r2-x2) dx = ... = (4/3). pi . r3
            /-r

Volume calculations

You can find area and volume calculations through this link

Length

We'll find a formula to calculate the length of a continuous curve y = f(x) in an interval [a,b].

In interval [a,b], we choose values x1,x2,x3,...,xn-1. Take a = x0 and b = xn.

In interval [xi-1,xi], we take the points P(xi-1,f(xi-1)) and Q(xi,f(xi)). We identify the curve in that interval, with the segment PQ.

 
   (length PQ)2  = (xi - xi-1)2 + (f(xi) - f(xi-1))2

                           (f(xi) - f(xi-1))2
                  = ( 1 + ------------------------) . (xi - xi-1)2
                             (xi - xi-1)2
According to Lagrange's theorem there is an si in [xi-1,xi], such that
 
     f(xi) - f(xi-1)
    --------------------- = f'(si)
      (xi - xi-1)
Now, we have
 
   (length PQ)2  = (1 + (f'(si))2 ) . (xi - xi-1)2

<=>
                    ________________
                   |
   (length PQ) =  \| 1 + (f'(si))2   . (xi - xi-1)
The sum of all the distances from each subinterval is a Riemann sum and the limit of this sum is a definite integral. The integral is the sum of an infinity number of elementary parts of the curve. Each elementary part has length
 
     _____________
    |
   \| 1 + (f'(x))2 dx

        _________
       |
   =  \| 1 + y'2  dx
Therefore we can write

The length of the curve y = f(x) in an interval [a,b] is
 
         /b    _________
         |    |
    L  = |   \| 1 + y'2  dx
         |
         /a

Example:

Calculate the length of the curve y = (ex + e-x)/2 in interval [-1,1].

 
    y' = (1/2).(ex - e-x)


  1 + y'2 = ... = (1/4).(ex + e-x)2


       /1
       |
   L = |  (1/2).(ex + e-x) dx  = ... = 2.35
       |
       /-1

Length calculations

You can find area, volume and length calculations through this link

Solved Problems

 
You can find solved problems about integration using this link
 




Topics and Problems

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