Let a curve y = F(x) be defined such that its gradient is given by dy dx = 24( ^2 x + ^2 x) ( x - x)^3 . If the curve passes through the point ( 8 , 1 ) , then the y -coordinate of the point on the curve where x = 12 is

Step-by-step solution

Given F(x) = 24( ^2 x + ^2 x) ( x - x)^3 dx . Let u = x - x . Then du = ( ^2 x + ^2 x) dx . Substituting this into the integral: F(x) = 24 u^3 du = 24 ( u⁻² -2 ) + C = - 12 u^2 + C F(x) = - 12 ( x - x)^2 + C We can simplify the denominator: x - x = x x - x x = ^2 x - ^2 x x x = - 2x 1 2 2x = -2 2x Thus, ( x - x)^2 = 4 ^2 2x . Substituting this back: F(x) = - 12 4 ^2 2x + C = -3 ^2 2x + C The curve passes through ( 8 , 1 ) , so F ( 8 ) = 1 : -3 ^2 (2 8 ) + C = 1 -3 ^2 ( 4 ) + C = 1 -3(1) + C = 1 C = 4 So, F(x) = -3