**Line Integral of a Vector Field: Lang (5.2.1)**

Here’s a problem (5.2.1, p. 58) from Lang’s “A Second Course in Calculus”:

Compute the line integral of the vector field over the indicated curve. *F*(*x, y*) = (*x*^{2} – 2*xy*, *y*^{2}-2*xy*) along the parabola *y* = *x*^{2} from (-2, 4) to (1,1).

To solve this problem, begin by finding a parametrization of the parabola *y *= *x*^{2} from (-2,4) to (1,1). We can do this by letting *x* = *t*, and then letting *y* = *t*^{2} for *t* in the interval [-2, 1]. Then, the parametrization is given by r(*t*) = (*t*, *t*^{2}).

We set *x* equal to the parameter *t *as we are moving down the parabola from *x* = -2 to *x* = 1. Then, because *y* = *x*^{2}, *y* = *t*^{2}.

Afterwards, we will find the derivative of the parametrization with respect to *t*: *r*’(*t*) = (1, 2*t*)

Of course, the derivative of *t* is merely 1, and the derivative of *t*^{2} is 2*t.*

Now, we will substitute *t* for *x* and *t*^{2} for *y* in *F*(*x, y*) = (*x*^{2} – 2*xy*, *y*^{2}-2*xy*).

*F*(*x, y*) = (*x*^{2} – 2*xy*, *y*^{2}-2*xy*)

*F*(*x*, *y*) = *F*(*r*(*t*)) = (*t*^{2} – 2*t*^{3}, *t*^{4} - 2*t*^{3})

Next, we have to take the dot product of the vector field along the curve with the derivative of the parametrization:

*F*(*r*(*t*)) * *r*’(*t*)

(*t*^{2} – 2*t*^{3}, *t*^{4} - 2*t*^{3}) * (1, 2*t*)

Using the definition of the dot product, we get:

(*t*^{2} – 2*t*^{3})(1) + (*t*^{4} - 2*t*^{3})(2*t*)

*t*^{2} – 2*t*^{3} + 2*t*^{5} – 4*t*^{4}

We can now compute the integral of this dot product over the interval [-2, 1].

Therefore, the answer is -36.9.

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