diketahui bahwa F'(x) = ax + b dan F(1)-F(0) = 4 ; F(2) -F (1) = 6. Dengan demikian nilai a.b =... Terima kasih..

F'(x) = ax + b
maka
F(x) = ½ax² + bx + c
F(1) - F(0) = 4
½a(1)² + b(1) + c - (½a(0) + b(0) + c) = 4
½a + b + c - (0 + 0 + c) = 4
½a + b = 4
F(2) - F(1) = 6
½a(2)² + b(2) + c - (½a(1)² + b(1) + c) = 6
½a(4) + 2b + c - (½a + b + c) = 6
2a + 2b + c - ½a - b - c = 6
1½a + b = 6
eliminasi dengan persamaan pertama
1½a + b = 6
½a + b = 4
----------------- _
a = 2
½a + b = 4
½(2) + b = 4
1 + b = 4
b = 3
a.b = 2.3 = 6