In this exercise we have to perform the parameterization of the given function and we will have:
[tex]\int\limits_C {f} \, dr = f(1, 2)-f(-1, 2)= 2/3[/tex]
In there is some scalar function [tex]f(x, y)[/tex] such that:
[tex]\nabla f(x,y) = f(x, y) = x^2i+y^2j[/tex]
Then we want to find [tex]f[/tex] such that:
[tex]f'(x)= x^2= f(x,y) = \frac{x^3}{3} + g(y)\\f'(y)= y^2= f(x,y)= \frac{y^3}{3} + C\\f(x, y)= \frac{x^3}{3} +\frac{y^3}{3} + C[/tex]
So the vector filed [tex]f(x, y)[/tex] is conservative, which means the fundamental theorem applies; the line integral of [tex]f[/tex] along any path [tex]C[/tex] parameterized by some vector-valued function [tex]r(t)[/tex] over [tex]a\leq t\leq b[/tex] is given by:
[tex]\int\limits_C {f} \, dr\\\int\limits^t=a_t=b {f(r(t))} \, dt= f(r(b))-f(r(a))[/tex]
In the case:
[tex]\int\limits_C {f} \, dr = f(1, 2)-f(-1, 2)= 2/3[/tex]
Learn more: brainly.com/question/14770282
