How do you set up the line integral $\int_{\gamma} \alpha$?

Question

Let $\alpha = dx_1 - x_1^{-1} dx_2$ and we have $\gamma(t) = (t^2, t^3)$, where $t$ is between $0$ and $1$.

Are we going to do something like this

$$\gamma'(t) = 2t d1 + 3t^2 d2.$$

Then $\alpha(\gamma') = 2t - 3x_1t^2$ and then we can plug in $t^2$ for $x_1$ and get

$$2t - 3t^4$$ and then integrate that from $0$ to $1$?

How would you confirm or deny the fact that you could solve this using the end points alone (line integrals)?

Answer 1

You're very close. To clean up and clear up notation I will start from the beginning. The differential 1-form $\alpha$ is notation for the following

$$\int_{\gamma} P(x_1,x_2)\; dx_1 + Q(x_1,x_2) \; dx_2=\int_{\gamma} P(x_1,x_2)\; dx_1 + \int_{\gamma} Q(x_1,x_2)\; dx_2$$

which can be evaluated over a parameterization $\gamma(t)$ as

$$\int_{\gamma(t)} P\big(x_1(t),x_2(t)\big)\; \frac{dx_1}{dt}\;dt + \int_{\gamma(t)} Q\big(x_1(t),x_2(t)\big)\; \frac{dx_2}{dt}\; dt \;\; .$$

We can clearly combine this into one integral involving $t$ and for your case this ends up being

$$\int_0^1 (1)(2t) - (t^{-2})(3t^2) \; dt = \int_0^1 2t - 3 \; dt \;\; .$$

Aside from some inconvenient notational choices, your only error was that you used $\alpha = dx_1 - x_1 dx_2$ instead of $\alpha = dx_1 - x_1^{-1} dx_2$ as the problem statement dictates.

Answer 2

How would you confirm or deny the fact that you could solve this using the end points alone (line integrals)?

Common confusion. You've plugged in the equation of curve into the one form. As you change the parameter of $t$, you are essentially moving along the curve $\gamma(t) = (t^2,t^3)$, so when you plug in the bounds of the parameter, you are actually taking the final coordinates on the curve.

So, at $t=1$, the point one curve is: $(1,1)$ And at $t=0$ , it is at $(0,0)$

So the curve is in such a way that it starts at the origin for the initial value of the parameter and ends at $(1,1)$ in the final value.