please tolong dong susah nih cuma nomer 1a doang bingung

Penyelesaian:

y = x²
y = x

y = y
x² = x
x² - x = 0
x(x - 1) = 0
x = 0 atau x = 1  (batas nilai x)

[tex] \int\limits^1_0 {(x) - (x^2)} \, dx [/tex]
= [tex] \int\limits^1_0 {-x^2 + x} \, dx [/tex]
= -1/3.x^3 + 1/2.x^2 ] 0->1
= -1/3(1)³ + 1/2(1)²  - (0)
= -1/3 + 1/2 = (-2+3)/6 
= 1/6 satuan luas