Area Using Integration Solution1





In this page area using integration solution1 we are going to see solution of some practice questions.

(1) Find the area of the region bounded by the line x - y = 1 and

(i) x - axis, x = 2 and x = 4

Solution:

Here we need to find the area of the bounded by the line x - y = 1 and x-axis, x = 2 and x = 4. By graphing the given line we come to know that the required area is above the x-axis. So we have to use the formula below.

                      b

Required area = ∫ y dx

                      a

                            4

                   = ∫ (x - 1)  dx

                      2

                                                4

                   = [(x²/2 - x)] 

                                      2

                   = (16/2 - 4) - (4/2 - 2)

                   =  (8 - 4) - (2 - 2)

                   =  4 - 0

                  =  4 square units 


(ii) x - axis , x = - 2 and x = 0

Solution:

Here we need to find the area of the bounded by the line x - y = 1 and x-axis, x = 2 and x = 4. By graphing the given line we come to know that the required area is below the x-axis. So we have to use the formula below.

                       b

Required area = ∫ - y dx

                      a

                            0

                   = ∫ (x - 1)  dx

                     -2

                                                -2

                   = [(x²/2 - x)] 

                                      0

                   = (4/2 - (-2)) - (0/2 - 0)

                   =  (2 + 2) - (0 - 0)

                   =  4 - 0

                   =  4 square units 


(2) Find the area of the region by the line x - 2 y - 12 = 0 and

(i) y - axis , y = 2 and y = 5

Solution:

                       b

Required area = ∫ x dy

                      a

                            5

                   = ∫ (2y + 12)  dy

                      2

                                             5

                   = [(2y²/2 + 12 y)] 

                                            2

                                        5

                   = [(y² + 12 y)] 

                                       2

                   = [5² + 12 (5)] - [2² + 12(2)]

                   =  [25 + 60] - [4 + 24]

                   =  85 - 28

                   =  57 square units 


(ii) y - axis , y = -1 and y = -3

                       b

Required area = ∫ -x dy

                      a

                            -3

                   = ∫ (2y + 12)  dy

                      -1

                                             -1

                   = [(2y²/2 + 12 y)] 

                                            -3

                                        -1

                   = [(y² + 12 y)] 

                                       -3

                   = [(-1)² + 12 (-1)] - [(-3)² + 12(-3)]

                   =  [1 - 12] - [9 - 36]

                   =  -11 - (-27)

                   =  -11 + 27

                   =  16 square units 

area using integration solution1 area using integration solution1