6. If (dy)/(dx)=4x^3-4x and y=0 when x=2 find y in terms of x. 7. At the point (x,y) on a curve, the gradient is equal to (2x-1) If the curve passes through the point (3,4) , what is the equation of the curve? 8. Find-the equation of the curve, gradient 2x^2+3x-1 , which passes through the origin. 19. If (dy)/(dx)=px+2 , where p is a constant, express y as a function of x.given that (d^2y)/(dx^2)=6 and that y=4 when x=0

6. To find y in terms of x, we need to integrate the given derivative $\frac{dy}{dx} = 4x^3 - 4x$ with respect to x. <br /><br />Integrating, we get:<br />$y = \int (4x^3 - 4x) dx = x^4 - 2x^2 + C$<br /><br />We are given that $y = 0$ when $x = 2$. Substituting these values into the equation, we can solve for the constant C:<br />$0 = 2^4 - 2(2^2) + C$<br />$0 = 16 - 8 + C$<br />$C = -8$<br /><br />Therefore, the equation of the curve is $y = x^4 - 2x^2 - 8$.<br /><br />7. We are given that the gradient of the curve at any point $(x, y)$ is equal to $2x - 1$. We can use this information to find the equation of the curve.<br /><br />Let's differentiate the equation of the curve with respect to x:<br />$\frac{dy}{dx} = 2x - 1$<br /><br />We are given that the curve passes through the point $(3, 4)$. Substituting these values into the equation, we can solve for the constant of integration:<br />$4 = 2(3) - 1 + C$<br />$4 = 6 - 1 + C$<br />$C = -1$<br /><br />Therefore, the equation of the curve is $y = x^2 - x - 1$.<br /><br />8. We are given that the gradient of the curve is $2x^2 + 3x - 1$, and it passes through the origin $(0, 0)$.<br /><br />To find the equation of the curve, we need to integrate the given gradient with respect to x:<br />$\int (2x^2 + 3x - 1) dx = \int dx$<br /><br />Integrating, we get:<br />$y = \frac{2}{3}x^3 + \frac{3}{2}x^2 - x + C$<br /><br />Since the curve passes through the origin, we can substitute $x = 0$ and $y = 0$ to solve for the constant C:<br />$0 = \frac{2}{3}(0)^3 + \frac{3}{2}(0)^2 - 0 + C$<br />$C = 0$<br /><br />Therefore, the equation of the curve is $y = \frac{2}{3}x^3 + \frac{3}{2}x^2 - x$.<br /><br />19. We are given that $\frac{dy}{dx} = px + 2$, where p is a constant, and $\frac{d^2y}{dx^2} = 6$. We are also given that $y = 4$ when $x = 0$.<br /><br />First, let's find the equation of the curve by integrating $\frac{dy}{dx}$ with respect to x:<br />$y = \int (px + 2) dx = \frac{p}{2}x^2 + 2x + C$<br /><br />We are given that $y = 4$ when $x = 0$. Substituting these values into the equation, we can solve for the constant C:<br />$4 = \frac{p}{2}(0)^2 + 2(0) + C$<br />$C = 4$<br /><br />Now, let's differentiate the equation of the curve with respect to x to find the second derivative:<br />$\frac{d^2y}{dx^2} = p$<br /><br />We are given that $\frac{d^2y}{dx^2} = 6$. Equating this to p, we get:<br />$p = 6$<br /><br />Therefore, the equation of the curve is $y = 3x^2 + 2x + 4$.