JAMB - Mathematics (2002 - No. 7)
If dy/dx = 2x - 3 and y = 3 when x = 0, find y in terms of x.
2x2 - 3x
x2 - 3x
x2 - 3x - 3
x2 - 3x + 3
Selitys
dy/dx = 2x - 3
y = ∫2x - 3 dx => y = x2 - 3x + c
3 = (0)2 - 3(0) + c
=> c = 3
Thus y = x2 - 3x + 3
y = ∫2x - 3 dx => y = x2 - 3x + c
3 = (0)2 - 3(0) + c
=> c = 3
Thus y = x2 - 3x + 3