Area Using Integration Solution2





In this page area using integration solution2 we are going to see solution of some practice questions.

(3) Find the area of the region bounded by the line y = x - 5 and the x - axis between the ordinates x = 3 and x = 7.

Solution:

Here we need to find the area of the bounded by the line y = x - 5 and x-axis, x = 3 and x = 7. By graphing the given line we come to know that the required area has divided into two parts.So we have to find the area between x = 3 and x = 5 which is below the x-axis and we have to find the area between x=5 and x = 7 which is above the x-axis and we have to add them.

                        b

Required area = ∫ y dx

                       a

                            5                     7

                   = ∫ (x - 5)  dx + ∫ -(x - 5)  dx

                      3                         5

                                                  5                     7

                   = [(x²/2 - 5x)]  +  [5x - x²/2 ]

                                       3                           5

                   = [5²/2 - 5 (5)] - [3²/2 - 5(3)] + [5(7) - 7²/2] - [5(5) - 5²/2]

                   = [25/2 - 25] - [9/2 - 15] + [35 - 49/2] - [25 - 25/2]

                   = [(25- 50)/2] - [(9-30)/2] + [(70 - 49)/2] - [(50 - 25)/2]

                   = (-25/2) + (21/2) + (21/2) - (25/2)

                   = (-25 + 21 + 21 - 25)/2

                   = (-50 + 42)/2

                   = 8/2

                  =  4 square units 


(4) Find the area of the region bounded by the curve y = 3 x² - x and the x - axis between x = -1 and x = 1.

Solution:

Here we need to find the area of the bounded by the the curve y = 3 x² - x and the x - axis between x = -1 and x = 1. This curve must be the equation of parabola.   area using integration solution2

To find the total area we have to area of three parts separately and we have to add them.  area using integration solution2

                        b

Required area = ∫ y dx

                       a

                            0                     1/3                   1

                   = ∫ (3 x² - x) dx + ∫ -(3 x² - x) dx + ∫ (3 x² - x) dx

                     -1                         0                           1/3

                            0                        1/3                   1

                   = ∫ (3 x² - x) dx + ∫ (x -3 x²) dx + ∫ (3 x² - x) dx

                     -1                              0                           1/3

                                                        0                        1/3                               1

                   = [(3x³/3 - x²/2)]  +  [x²/2 - 3x³/3] +  [(3x³/3 - x²/2)]

                                          -1                                0                                 1/3

                                                 0                   1/3                        1

                   = [(x³ - x²/2)]  +  [x²/2 - x³] +  [(x³ - x²/2)]

                                      -1                         0                           1/3

                       = [0 - (-1-(1/2))]+[(1/18)-(1/27)-0]+[(1-1/2)-(1/27-1/18)]

                       = [-(-3/2)]+[(3-2)/54]+[(2-1)/2-(2-3)/54]

                       = [3/2]+[1/54]+[(1/2)-(-1/54)]

                       = [81/54]+[1/54]+[(27+1)/54]

                       = (81 + 1 + 28)/54

                       = 110/54

                  = 55/27 square units

area using integration solution1 area using integration solution1