Find the equation of the curve for which y'=3x² and which passes through (1,-1).
To find the equation of the curve, we need to integrate the given derivative equation with respect to x and then solve for the integration constant using the given point.
First, let's integrate the derivative equation y' = 3x^2 with respect to x:
∫dy/dx dx = ∫3x^2 dx
Integrating both sides gives:
y = ∫3x^2 dx
To find the antiderivative of 3x^2, we can use the power rule of integration:
y = x^3 + C
where C is the integration constant.
Now, we have the equation of the curve before applying the initial condition.
Next, we can use the given point (1, -1) to find the specific value of the integration constant C.
Substituting x = 1 and y = -1 into the equation y = x^3 + C:
-1 = 1^3 + C
Simplifying this equation:
-1 = 1 + C
C = -1 - 1
C = -2
Now that we have the value of the integration constant C, we can replace it in the equation of the curve:
y = x^3 - 2
Therefore, the equation of the curve that satisfies y' = 3x^2 and passes through (1, -1) is y = x^3 - 2.