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 j

r(u) = u i + 2u^2 j

0 ? u ? 1

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.

The value of the line integral is 2

Step-by-step explanation:

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

With a(y) a function that depends only on y. A general primitive of yx² j (over y) is

With b(x) only depending on x

The 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

note that

As we wanted. Lets find the endpoints of C

r(u) = u i + 2 u² j

r(0) = (0,0)

f(1) = (1,2)

Therefore,

The value of the line integral over C is 2.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b