Find the curve whose slope at the point (x, y) is 3x^2 if the curve has to pass through the point (1,-1)?

The slope of the curve at (x , y)
=3x^2.

Therefore the curve must be the premtive of dy/dx =
3x^2.

So the curve is y = Integral
(3x^2)dx

y = (3x^3)/4+ c, Or y = x^3+c ,where  c is a
constant. Since this curve passes through (1 , -1), the coordinates (1 , -1) should
satisfy y = x^3+c.

-1 =
1^3+c.

c = -1-1 = -2

Therfore
the required curve is y = x^3 -2.