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My Teacher gave me this question.

Let C be the Curve $$y=x^3-x, \text{ where: } 0< x< 1$$ Oriented from $(0,0)$ to $(1,0)$. Find: $$\int_c x \, dy$$ I want to use the Line-Integral Formula: $$\int_c f(x,y)= \int_a^b f(h(t),g(t)) \cdot (\text{magnitude})r'(t) \, dt$$

My Parametrizaiton is: $x=t;y=t^3-t$, and my $r'(t) =\langle 1,3t^2-1\rangle, (\text{magnitude})r'(t)=\sqrt{1^2+(3t^2-1)^2} $

$(\text{magnitude})r'(t)$ is needs to be a number , but what numbers am I supposed to plug in to the $r'(t)$ to find my magnitude?. If i plug in $1$, I get $\sqrt5$ if I plug in $0$, I get $\sqrt 2$.

I tried to solve it by ignoring path length all together, just cutting $r'(t)$ out of the line integral equation, and I got the correct answer $(1/4)$.

So either the path length is irrelevent? maybe b/c the starting point is $(0,0) \to (1,0)$ and the length is one even though the function is a curve? or the path length is 1? Can someone please help explain this to me?

Michael Hardy
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Jess L
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1 Answers1

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There is a theorem that integrals are not path dependent if your vector field (in your case this is $dy$) is the derivative of a function. For you, this function is exactly $y(x)=x^3 - x$, so yes, your line integral will be path independent for this vector field.

Suppose we are evaluating a line integral along a parametrized curve: $\phi :[a,b]\rightarrow \mathbb{R}^n$, in your case, $\mathbb{R}^3$. Let the vector field with respect to which our line integral is being evaluated be $\nabla F$ for some potential funciton $F$.

Then $\int_{\phi}\nabla F dt= \int_{a}^{b} \nabla F(\phi(t))\cdot \phi'(t) dt = F(\phi(b))-F(\phi(a))$ by the fundamental theorem of line integrals and chain rule (you should verify this for yourself).

As you can see, our potential function will give the same thing no matter what path we take as it is given by the difference of our potential function evaluated at the two points.

operatorerror
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