Solve the intial value problem using integrals: [tex]dy/dx[/tex] = (cube root of x) , y(1) = 2 The solution: if f(x) = F(x), and y(x) is equal to F(x), then, y(x) = ([tex]x sub 0[/tex], [tex]y sub 0[/tex]) Here's what I did, and failed: y(1) = 2. This simply means (and I understand this) that if x is 1, then the function y(x) is 2 from the input of x equal to 1. So, in order to find the slope of the curve at the cube root of x, when y(1) = 2. We must integrate. this is the problem I am having. I get: y = [tex]integral[/tex] cube root of x [tex]dx[/tex] + C [tex]dy/dx[/tex] = cube root of 1 (substitued the intial value of x) + C At the coordinates (1,2) What am I doing wrong here? I dont' know how to integral the cube root of 3. Amd what is C? 2? Thank you.