91Ó°ÊÓ

A parabola is a U-shaped curve that is defined mathematically as the set of all points that are equidistant from a fixed point called the focus and a line known as the directrix. In the context of the given problem, our equation is \( y^2 = 2x \). This is a form of a parabola that opens to the right, with its vertex at the origin \((0,0)\).
When dealing with parabolas, it is important to consider:

• The vertex, which is the point where the curve changes direction. For \( y^2 = 2x \), the vertex is at the origin.


• The direction in which it opens. A parabola opening to the right like \( y^2 = 2x \) suggests that \( x \) increases as \( y \) increases both positively and negatively.


• The axis of symmetry, a line that divides the parabola into mirrored halves. For \( y^2 = 2x \), the line \( x = 0 \) or the y-axis is the axis of symmetry.

Understanding these properties helps in visualizing how the parabola interacts with other equations or constraints, such as the linear equation \( y = 4x - 1 \) in the problem.

Quadratic equations are polynomial equations of degree 2, typically in the form \( ax^2 + bx + c = 0 \). In our problem, after substituting \( y = 4x - 1 \) into \( y^2 = 2x \), we derived the quadratic equation:
\[16x^2 - 10x + 1 = 0\]
Quadratic equations can be solved using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a = 16 \), \( b = -10 \), and \( c = 1 \). The discriminant \( b^2 - 4ac \) determines the nature of the roots of the quadratic equation. In this case, it is \( 36 \), which is positive, indicating two distinct natural roots.
Solving gives the roots at \( x = \frac{1}{2} \) and \( x = \frac{1}{8} \), the x-coordinates of our intersection points. These roots guide us in setting the bounds of integration needed to find the area of the region.

In calculus, a definite integral represents the area under a curve between two specified points along the x-axis. For our exercise, we focus on finding the area between the parabola \( y^2 = 2x \) and the line \( y = 4x - 1 \) from \( x = \frac{1}{8} \) to \( x = \frac{1}{2} \).
The integral setup is:
\[ \int_{\frac{1}{8}}^{\frac{1}{2}} (\sqrt{2x} - (4x - 1)) \, dx \]
Here, \( \sqrt{2x} \) represents the upper curve (parabola) and \( 4x - 1 \) the lower line. Calculating the integral involves finding:

• The antiderivative of \( \sqrt{2x} \), which is \( \sqrt{2} \cdot \frac{2}{3}x^{3/2} \).


• The antiderivative of \( -4x \), which is \( -2x^2 \).


• The antiderivative of \( 1 \), which is \( x \).

After integration, we substitute the bounds \( x = \frac{1}{8} \) and \( x = \frac{1}{2} \) to find the specific area between these curves, yielding a result of \( \frac{5}{64} \) square units.