91Ó°ÊÓ

When dealing with problems in integral calculus involving the shape of a semicircle, it's crucial to understand the basics of a circle's geometry. A full circle's equation is typically given by \(x^2 + y^2 = r^2\), where \(r\) is the radius.
For this exercise, we look at the equation \(x^2 + y^2 = 9\), representing a circle with a radius of 3. The function \(\sqrt{9-x^2}\) gives us the upper semicircle because it defines the positive \(y\)-values over the interval. This segment lies above the x-axis, providing us a semicircle with radius 3, extending from \(-3\) to \(3\).
To find the area of a full circle, we use \(\pi r^2\). Given our radius \(r = 3\), the area of a complete circle is \(9\pi\). However, since we are only interested in a semicircle, we take half of this area, resulting in \(\frac{9\pi}{2}\).

• The radius in this problem is 3.
• Full circle's area: \(9\pi\).
• Semicircle's area: \(\frac{9\pi}{2}\).

The concept of a constant function is foundational in calculus, particularly when computing areas under curves. A constant function is simply a horizontal line at a given height \(c\). Its integral over an interval \([a, b]\) is equivalent to the area of a rectangle with width \(b - a\) and height \(c\).
In the given problem, the function portion \(1\) is a constant function. It spans from \(-3\) to \(0\) on the x-axis. Thus, the area under this constant function is a rectangle with a base of 3 (since \(0 - (-3) = 3\)) and a height of 1.
The calculation of this area proceeds as follows:

• Width of the rectangle: 3
• Height of the rectangle: 1
• Area of the rectangle: 3 \(\times\) 1 = 3

A definite integral computes the accumulation of quantity, often interpreted as area under a curve over a specific interval. In our exercise, this is represented by \(\int_{-3}^{0}\left(1+\sqrt{9-x^{2}}\right) dx\), which combines elements of both the constant function and semicircle.
When evaluating this integral, understand it as the sum of two separate areas:

• The area under the constant function \(1\), over the interval \(-3\) to \(0\). This is a rectangle, as discussed previously (area = 3).
• The area of a semicircle segment from \(-3\) to \(0\), which is exactly half of the semicircle area (\(\frac{9\pi}{4}\)).

Together, these areas give the integral's value. Adding them up, the definite integral equals \(3 + \frac{9\pi}{4}\).
A few key points about definite integrals:

• They are used to find total areas accumulated.
• They can represent physical quantities like distance or volume.
• They allow us to consider areas below the x-axis as negative, though not applicable here.