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Completing the square is a helpful technique in algebra used to simplify quadratic expressions.
This process involves expressing a quadratic expression like \( ax^2 + bx + c \) in the form \( a(x-h)^2 + k \). To complete the square for the expression \( x^2 - 6x \):

• Identify the coefficient of \( x \), which is \(-6\).
• Take half of \(-6\), which is \(-3\), and then square it to get 9.
• Add and subtract this square within the expression: \( x^2 - 6x = (x-3)^2 - 9 \).

So, the expression \( 6x - x^2 \) becomes \( 9 - (x-3)^2 \), allowing you to simplify by taking the square root to \( \sqrt{9-(x-3)^2} \).
This transformation is crucial as it reveals the geometric representation inside the integral.

A semicircle is half of a circle and has unique properties in geometry.
The general equation of a semicircle derived from a circle centered on the x-axis is \( y = \sqrt{r^2 - (x-h)^2} \), where \( r \) is the radius of the circle, and \( h \) is the horizontal shift.
In our integral, the function \( \sqrt{9-(x-3)^2} \) describes a semicircle with:

• Radius \( r = 3 \).
• Center at \( (3,0) \).

This mathematical form matches a semicircle's upper half. Visually, it looks like a hill or an arc between its starting point and endpoint.
The semicircle's curve is symmetric about the line \( x = 3 \), creating a perfect half-circle representation above the x-axis.

The area of a circle is a fundamental concept in geometry, calculated with the formula \( \pi r^2 \). For a full circle with radius \( r \), this represents the total space enclosed by its circumference. However, when dealing with semicircles or segments of a circle, the area needs to be adjusted accordingly.
For instance, a semicircle's area is half of a full circle, calculated as \( \frac{1}{2} \pi r^2 \). When evaluating areas under specific integral bounds, such as \( x = 0 \) to \( x = 3 \) in our context:

• This represents a quarter segment of a semicircle because it covers only a quarter of the circle's total 360-degree rotation.

For a circle with radius 3, the full circle's area is \( 9\pi \) (since \( 3^2 = 9 \)). Thus, the area under the given integral becomes a calculation of \( \frac{1}{4} \times \frac{1}{2} \times 9\pi = \frac{9\pi}{8} \), representing the desired quarter-segment above the x-axis.