Question

Calculate \(\int_{a}^{b} f(x) d x\), where a and \(b\) are the left and right end points for which f is defined, by using the Interval Additive Property and the appropriate area formulas from plane geometry. Begin by graphing the given function. $$ f(x)=\sqrt{A^{2}-x^{2}} ;-A \leq x \leq A $$

Short answer

The integral is \(\frac{\pi A^2}{2}\), the area of the semi-circle.

Step-by-step solution

Graph the Function

The function \(f(x) = \sqrt{A^2 - x^2}\) represents a semi-circle with radius \(A\) centered at the origin. Plot this semi-circle spanning from \(-A\) to \(A\) along the x-axis as the semi-circle is part of the upper half plane.

Recognize the Geometrical Shape

The function represents the upper half of a circle with radius \(A\). Hence, our region of interest is a semi-circle.

Use the Area Formula for a Circle

The area of a circle is \(\pi r^2\). Since the semi-circle's radius is \(A\), its full circle area would be \(\pi A^2\). The semi-circle, our region of interest, occupies half of this area, giving us \(\frac{\pi A^2}{2}\).

Apply the Interval Additive Property

The Interval Additive Property tells us that the integral can be split over intervals. However, in this case, since we are considering the whole semi-circle, we don't need to split intervals. We'll apply the integral for the whole potential domain \([-A, A]\).

Calculate the Integral

The area under \(f(x) = \sqrt{A^2 - x^2}\) from \(-A\) to \(A\) is equivalent to the area of the semi-circle. Therefore, \(\int_{-A}^{A} \sqrt{A^2 - x^2} \, dx = \frac{\pi A^2}{2}\).

Key concepts

Semi-Circle Integration

Integrating a function like \( f(x) = \sqrt{A^2 - x^2} \) involves understanding the shape it represents. In this case, it describes a semi-circle. This might sound complex, but let's break it down. The function \( \sqrt{A^2 - x^2} \) is derived from the equation of a circle \( x^2 + y^2 = A^2 \).
When solving for \( y \), we obtain two results: \( y = \sqrt{A^2 - x^2} \) and \( y = -\sqrt{A^2 - x^2} \). These represent the upper and lower halves of a circle, respectively. By focusing on the function \( f(x) = \sqrt{A^2 - x^2} \) for the interval \(-A \leq x \leq A\), we essentially investigate the upper half.
Thus, the task of integrating this function simplifies to determining the area beneath this semi-circle. Rather than directly calculating via integration, we can leverage the known geometric area of this shape.

Interval Additive Property

The Interval Additive Property is a useful concept in integration. It states that the integral over an interval can be broken into several parts, or summed, over sub-intervals. This is especially helpful when dealing with functions that have complex or discontinuous domains.
For the function \( \sqrt{A^2 - x^2} \), the entire interval from \(-A\) to \(A\) is continuous and smooth because it is a semi-circle. Thus, there is no real need to split the interval.
However, understanding this property helps recognize that any integration from a starting point \(a\) to an endpoint \(b\), can be divided to make calculations simpler in more complex scenarios. The key takeaway is to realize that while you can split intervals at will, knowing when it’s not needed is equally essential.

Geometric Area Formula

A useful trick in integration, particularly when dealing with shapes like circles and semi-circles, involves applying geometric area formulas. The familiar formula for the area of a circle is \( \pi r^2 \). For a semi-circle, which is precisely half of a full circle, the area is simply half of \( \pi r^2 \), or \( \frac{\pi r^2}{2} \).
In this exercise, since \( f(x) \) creates an upper half-circle, we only need the semi-circle area. The radius here is \( A \), yielding the area \( \frac{\pi A^2}{2} \). Using this method circumvents the need for more tedious integration calculations, as this problem uses geometric understanding to arrive directly at the result.

• Recognizing shapes simplifies problem-solving drastically.
• Directly using area provides a shortcut to solving definite integrals of simple geometric shapes.

By applying such geometrical insights, integrations tied to basic shapes become direct and visual-friendly.