The slope of the tangent to a curve y = f(x) at any point (x, y) is given by dy dx = x^3 x^2 - 1 . If the curve passes through the point ( 2 , 1) , then the y -coordinate of the point on the curve where x = 5 is equal to

Step-by-step solution

Given dy dx = x^3 x^2 - 1 Integrating both sides with respect to x , we get: y = x^3 x^2 - 1 dx Let x^2 - 1 = t^2 2x dx = 2t dt x dx = t dt Also, x^2 = t^2 + 1 Substituting these into the integral: y = (t^2 + 1) t t dt y = (t^4 + t^2) dt y = t^5 5 + t^3 3 + C Since t = x^2 - 1 , the equation of the curve is: y = (x^2 - 1)^ 5/2 5 + (x^2 - 1)^ 3/2 3 + C The curve passes through ( 2 , 1) . Substituting x = 2 and y = 1 : 1 = (2 - 1)^ 5/2 5 + (2 - 1)^ 3/2 3 + C 1 = 1 5 + 1 3 + C 1 = 8 15 + C C = 7 15 So, the curve is y