91Ó°ÊÓ

A parabola is a symmetrical, U-shaped curve that can open either upwards or downwards on a graph. The equation of a parabola in its simplest form is given by \( y = ax^2 + bx + c \). In our problem, the parabola is described by the equation \( y = x^2 \), which means our parabola opens upwards, starting from the origin point \((0,0)\).
- It has a vertex, which is the lowest point for an upward-opening parabola.- The vertex of the parabola \( y = x^2 \) is at \( (0, 0) \).- A parabola is symmetric along its vertical line of symmetry, which, for \( y = x^2 \), is the y-axis.
Understanding parabolas helps in recognizing the curve shapes in graphical problems, which is crucial for calculating specific types of integrals like line integrals.

Parameterization is the process of expressing a curve using a single parameter, often denoted by \( t \). In line integrals, this helps by taking multi-dimensional curves and simplifying them into single-variable expressions.
For the equation \( y = x^2 \), to parameterize the curve, assign:

• \( x = t \)
• \( y = t^2 \)

Here, \( t \) varies from the starting point of the curve \((t = 0)\) to the endpoint \((t = 1)\).
Parameterization helps simplify the integral by transforming derivatives into terms of \( dt \), such as \( dx = dt \) and \( dy = 2t \cdot dt \). This simplification is valuable for easily substituting into the integral for evaluation.

Finding the antiderivative, also known as the indefinite integral, is a fundamental part of solving integrals. It involves determining a function whose derivative is the given function.
In the problem, after substituting the parameterized expressions into the integral, you need to find the antiderivative of \( 3t^2 \). The antiderivative of \( 3t^2 \) is found using the power rule:\[ \int 3t^2 \, dt = t^3 + C\]
Here, \( C \) is the constant of integration, which we typically set aside in definite integrals as we evaluate over a specific interval.
Knowing how to find antiderivatives allows you to handle a variety of integrals, which is an essential skill in calculus.

Integral evaluation is the process of computing the value of definite integrals over a specified interval. This involves using the antiderivative to find values at boundary points and subtracting them.
For our problem, compute the definite integral:\[ \int_{0}^{1} 3t^2 \, dt\]Using the antiderivative \( t^3 \), evaluate from \( 0 \) to \( 1 \):\[ \left[ t^3 \right]_{0}^{1} = (1)^3 - (0)^3 = 1\]
The computed value, \( 1 \), gives the result of the line integral over the curve \( y = x^2 \) from \((0,0)\) to \((1,1)\). Integral evaluation is crucial for determining the total "weight" or "sum" of a function over a path or area.