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When we are faced with evaluating line integrals over curves, the first step is often to parametrize the curve. Parametrization refers to representing a curve using a parameter, typically denoted as t, which when varied within a certain interval, describes the entire curve in a space.

For example, the given curve in our problem is y = x², and we want to express it using a parameter t. Setting x=t and y=t² defines a new way to represent points on the curve where t varies from 0 to 1. The resulting parametric equations x(t) = t and y(t) = t² make up the vector-valued function \(\boldsymbol{r}(t) = \langle t, t^2 \rangle\).

This new representation simplifies the process of integrating over the curve. By using parametrization, you can convert a complex curve in space into a simpler problem of dealing with a single variable t, which eventually makes the integration process more straightforward.

Arc length is a crucial concept when dealing with scalar line integrals as it embodies the notion of 'distance' along a curve. It's used to find the length of a curved path and is symbolized by the differential ds. In our problem, we need to evaluate ds to integrate along the curve C.

For a parametrized curve \(\boldsymbol{r}(t) = \langle x(t), y(t) \rangle\), the arc length differential ds is the magnitude of the derivative of \(\boldsymbol{r}(t)\) with respect to the parameter t, abbreviated as \(||\frac{d\boldsymbol{r}}{dt}||\). Mathematically, for a two-dimensional curve with the parametrization above, ds is given by \(ds = \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } dt\).

Computing the arc length allows us to substitute the originally complicated path-dependent integral by an easier one with respect to the parameter t, which contributes essentially to the problem's simplification and solution.

The substitution method is a vital technique in integral calculus, commonly known as u-substitution. It simplifies the integration process by changing the variable of integration to a new variable u, which reduces the integrand to a simpler form.

In our exercise, we applied the substitution method by setting u = 1 + 4t², which transforms the integral into a new limit in terms of u instead of t. The crucial step here is to express dt as a function of du, which is done by differentiating u with respect to t to find du/dt, and then solving for dt.

After substitution, the integral generally becomes easier because you have a familiar form, which can be integrated using basic integration rules. This powerful technique often simplifies otherwise challenging integrals and is a frequent ally in solving various calculus problems.