Here’s a problem (5.2.2, p. 58) from Lang’s “A Second Course in Calculus”:

Compute the line integral of the vector field over the indicated curve. (*x*, *y*, *xz *- *y*) over the line segment from (0, 0, 0) to (1, 2, 4).

*x*(*t*) = *x*_{0 }+ (*x*_{1} – *x*_{0})*t*

*y*(*t*) = *y*_{0 }+ (*y*_{1} – *y*_{0})*t*

*z*(*t*) = *z*_{0 }+ (*z*_{1} – *z*_{0})*t*

Here, *t* varies between 0 and 1. When *t* is 0, you get (0, 0, 0); when *t* is 1, you get (1, 2, 4). Let’s get the parametrization now:

*x*(*t*) = 0 + (1 - 0)*t = t*

*y*(*t*) = 0 + (2 - 0)*t = *2*t*

*z*(*t*) = 0 + (4 - 0)*t *= 4*t*

Therefore,

*r*(*t*) = (*t*, 2*t*, 4*t*)

*r*’(*t*) = (1, 2, 4)

Now, we will substitute *t* for *x*, *2t *for *y*, and 4*t* for *z *in the vector field:

*F*(*r, t*) = (*x*, *y*, *xz *- *y*)

*F*(*r, t*) = (*t*, 2*t*, 4*t*^{2} – 2*t*)

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*t*, 4*t*^{2} – 2*t*) * (1, 2, 4)

Using the definition of the dot product, we get:

*t *+ 4*t* + (4*t*^{2} – 2*t*)(4)

5*t*+ 16*t*^{2} – 8*t*

16*t*^{2} – 3*t*

We can now compute the integral of this dot product over the interval [0, 1].

Therefore, the answer is 23/6.

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