My teacher said that this is not necessary in the line integral, but why? Question:
Calculate the Scalar line integral:
$$\int_C \left(\, − \,\right)$$
C is the segment traveled in the direction: $$(1,1)\,to\,(2,3)$$
I started by solving this question by parameterizing the points.
Parameterization formula: $$(x,y)=B\,t+(1-t)\,A$$
$$(x,y)=(2,3)\,t+(1-t)\,(1,1)$$
$$(x,y)=(2t,3t)+(1-t,1-t)$$
$$(x,y)=(1+t,1+2t)$$
Our parameterization determines that $$ 0 ≤ t≤1$$
So our function $$r(t)=(1+t)\hat{i}+(1+2t)\hat{j}$$
$$r'(t)=(1,2)$$
$$ |r'(t) |=\sqrt{1^{2}+2^{2}}=\sqrt{5}$$
So our scalar line integral will be
$$\int_C \, − \, =\int_0^1 [(1+t)\,1 - (1+2t)\,2 ]\,|r'(t) | dt$$
$$=\int_0^1 [1+t - 2-4t ]\,\sqrt{5}dt $$
$$=\sqrt{5} \int_0^1 -3t - 1 dt$$
$$\therefore\sqrt{5}\,(\frac{-5}{2})$$
The thing is, my professor said that it is not necessary to have put the
$$|r'(t) |$$
in the scalar integral formula, because that was another case, but he didn't explain... was he right?
 A: Your teacher is correct. I have always hated the notation $\int_{\gamma}f_{1}\,dx + f_{2}\,dy$ for line integrals because these involves the notion of differential forms which the readers of a first course in vector calclulus might not be familiar with.
I always use Murray Spiegel's notations which tries to relate the line integral with the work done by a force $f$ in moving a particle along the path $C$ .
Here's how I would write it . Define $f=x\hat{i}-y\hat{j}$.
And $\vec{dr}=dx\hat{i} + dy\hat{j}$
Then the line integral over the curve $C$ is given by
$$\int_{C} f\cdot \vec{dr}$$ . Note that here $\cdot$ refers to the dot product.
Now you parametrize the curve $C=\{(1+t,1+2t):t\in[0,1]\}$ .
Then $dx\hat{i}+dy\hat{j}=d(1+t)\hat{i}+d(1+2t)\hat{j}=\bigg(\hat{i}+2\hat{j}\bigg)\,dt$.
Then $$\int_{C}f\cdot\vec{dr}=\int_{0}^{1}\bigg((1+t)\hat{i}-(1+2t)\hat{j}\bigg)\cdot\bigg(\hat{i}+2\hat{j}\bigg)\,dt=\int_{0}^{1}\bigg((1+t)-2(1+2t)\bigg)\,dt$$.
Now when you do $\int_{C}f\,|r'(t)|dt$ what you are evaluating is $\int_{C}f\,\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}\,dt$ which is not the same at all. Recall that the work done by a force is defined as $\int f\cdot dS$ and not $\int f\cdot |dS| $.
So all in all remember that the Line integral of a vector field(force) $f$ is just the work done by the force in moving a particle along the curve $C$.
