Verify that the given vector field h is a gradient. Then calculate the line integral of h over the indicated curve C by finding f such that the gradient f = h and evaluating f at the end points of C.h(x,y) = xy^2 i + yx^2 jr(u) = u i + 2u^2 j0 ? u ? 1Please show all the steps so that I will know where i am making a mistake. I have attempted this question 4 times and i do not know where am going wrong. The correct answer is 2. I keep getting 4.

Answer:[tex] f(x,y) = \frac{x^2 y^2}{2} [/tex]The value of the line integral is 2Step-by-step explanation:[tex] h(x,y) = xy^2 i + yx^2 j [/tex] Note that if you derivate the first part over the variable y and the second part over the variable x, then in both cases you obtain 2xy, therefore there must be a function f whose gradient is h, because the cross derivates are equal. In order to find such f, you can calculate a primitive of both expressions, the first one over the variable x and the second one over the variable y.A general primitive of xy² i (over x) is [tex] f_1(x,y) = \frac{x^2y^2}{2} + a(y) [/tex]With a(y) a function that depends only on y. A general primitive of yx² j (over y) is [tex] f_2(x,y) = \frac{x^2y^2}{2} + b(x) [/tex]With b(x) only depending on xThe function f(x,y) whose gradient is h is obtained by equaling the expressions of f₁ and f₂. f₁ and f₂ are equal when a(x) = b(x) = 0, therefore [tex] f(x,y) = \frac{x^2y^2}{2} [/tex]note that fx(x,y) = xy²fy(x,y) = yx²As we wanted. Lets find the endpoints of Cr(u) = u i + 2 u² jr(0) = (0,0)f(1) = (1,2)Therefore, [tex]\int\limits_C {xy^2} \, dx + {yx^2} \, dy = f((1,2)) - f((0,0)) = \frac{1^2 * 2^2}{2} - \frac{0^20^2}{0} = \frac{4}{2} - 0 = 2[/tex]The value of the line integral over C is 2.