Integration
Basic Integration
      ∫ a dx=ax +c
                                 n +1
      ∫ ax n dx = an+1
                     x
                                        +c
                                   ( a+ bx )n+1
      ∫ ( a+bx )n dx= (n+1)×b +c                     (But this formula is only applicable if the function is linear)
                            ax
      ∫ eax dx= ea              +c
                             −cosax
      ∫ sinax dx=             a
                                    +c (Please remember this formula is only applicable if the highest
       power of sin and variable x is 1)
                                 sinax
      ∫ cosax dx=                 a
                                       +c (Please remember this formula is only applicable if the highest
       power of cos and variable x is 1)
       Note:
       If they ask to integrate something related to sin 2 x , si n3 x , cos2 x∨cos 3 x please
       remember that it can’t be done directly. Some derivations will be provided in previous
       parts of the question. You have to use those to substitute and make a function with
       highest trigonometric power 1 and then integrate.
Integration with limits
        5
       ∫ (x ¿¿ 2+ 3)dx ¿
        2
                 3      5
                x
       ¿[         +3 x ]
                3       2
       ¿    {
          ( 5 )3
            3
                 +3 ( 5 ) − }{
                            ( 2 )3
                              3
                                   +3 ( 2 )       }
       ¿ 48
                                Application of Integration Area
          When a curve or a line is integrated with respect to x within given limits. We
           find the area bounded by the curve or line and the x-axis.
          When a curve or a line is integrated with respect to y within given limits. We
           find the area bounded by the curve or line and the y-axis.
          While integration within limits, the limits has to be specifically chosen, keeping
           in mind that the curve is on the same side of the axis.
           That is below the axis portion has to separately integrated and above the axis
           separately and then the positive values for the areas have to be added.
          To find the area bound by only curve and line or only curve and curve
           Step1
           Find the coordinates of intersection points
           Step2
           Integrate within limits the function line minus curve or curve minus line or
           curve minus curve and the limits the limits would be the coordinates of
           intersection point. If value comes negative, take modulus.
                              Application of Integration Volume
          When a curve or line is rotated 360/2 π about the x-axis/y-axis a 3-D shape is
           formed.
          The volume of the shape is given by the formula
       b
V =π ∫ y dx
             2
       a
(This is when rotated about x-axis and a and b are points on x coordinates about
which it is being rotated.)
       b
V =π ∫ x dy
            2
       a
(This is when rotated about y-axis and a and b are points on y coordinates about
which it is being rotated.)
          To find the volume bound by only curve and line or only curve and curve
           rotated about x-axis
           Step1
           Find the coordinates of intersection points
           Step2
       b
V =π ∫ ( y ¿¿ 1 − y 2)dx ¿
                  2   2
       a
π times integration within limits the function
line square minus curve square
curve square minus line square
curve square minus curve square
and the limits the would be the x-coordinates of intersection point.
If value comes negative, take modulus.
     To find the volume bound by only curve and line or only curve and curve
        rotated about y-axis
        Step1
        Find the coordinates of intersection points
        Step2
      b
V =π ∫ (x ¿¿ 1 −x 2 )d y ¿
              2    2
      a
π times integration within limits the function
line square minus curve square
curve square minus line square
curve square minus curve square
and the limits the would be the y-coordinates of intersection point.
If value comes negative, take modulus.