Luas daerah yang dibatasi oleh kurva y = 2x^2 - 6x + 4 dan -1≤ x ≤ 2 adalah....

y = 2x² -6x + 4,  sumbu x , -1 ≤ x ≤ 2
2x² -6x + 4 =0
2(x² - 3x +2)=0
2(x -2)(x-1)= 0
x = 1, x = 2

L = ₋₁¹∫ 2x ² -6x + 4 dx - ₁²∫2x² - 6x + 4 dx
L= (2/3 x³ - 3x² + 4x]¹₋₁ - [2/3 x³ - 3x² + 4x ] ²₁

L = 2/3(1+1) - 3(1 -1) + 4(1 +1)  - [ 2/3(8-1) - 3(4-1)+ 4(2-1)]
L = 4/3 +8 - [ 14/3 - 9 + 4]
L = 4/3 + 8 - 14/3 + 9 - 4
L =  - 10/3 + 13 = 29/13