Problem 81 - Calculus Volume 1

Question

In the following exercises, evaluate the integral using area formulas. \(\int_{1}^{5} \sqrt{4-(x-3)^{2}} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(2\pi\).

Step 1: Recognize the Integral's Geometric Interpretation

Observe the integral \( \int_{1}^{5} \sqrt{4-(x-3)^{2}} \, dx \). This expression resembles the equation of a circle centered at \((3, 0)\) with a radius of 2, specifically the upper half of the circle, since we consider the positive square root.

Step 2: Identify the Circle's Position

The standard form of a circle \((x-a)^2 + y^2 = r^2\) translates this expression to \((x-3)^2 + y^2 = 4\). This tells us the center is \((3,0)\) and the radius is 2. We consider only \(y = \sqrt{4 - (x-3)^2}\), which is the upper half of the circle.

Step 3: Determine the Area Covered by Limits

The integration limits \(x = 1\) to \(x = 5\) indicate that we take the segment of the semicircle from \(x = 1\) to \(x = 5\). The whole semicircle extends from \(x = 1\) to \(x = 5\), as 1 to 5 is symmetric around the center at \(x = 3\).

Step 4: Calculate Area Using the Semicircle Formula

The semicircle has a radius of 2, thus its area is half of a full circle with \( r = 2 \). The area of a full circle is \( \pi r^2 = \pi (2)^2 = 4\pi \). Therefore, the area of the semicircle is \( \frac{1}{2} \times 4\pi = 2\pi \).

Step 5: Conclude the Integral Evaluation

Since the integral represents the area under the curve from \(x = 1\) to \(x = 5\) for the upper semicircle of the circle, the value of the integral is equal to the area of the semicircle, which is \(2\pi\).

Key Concepts

Geometric Interpretation of Integrals

When working with definite integrals, it's helpful to understand their geometric meaning. This integral allows us to compute the area under a specific curve. In this example, the curve forms part of a semicircle. Integrals are viewed as the sum of many small slices or rectangles under the curve across the range specified by the limits of integration. Here, the limits from 1 to 5 define the width of the area of interest.

Visualizing the geometric shape associated with the integral can simplify the process of integration. Instead of solving analytically, we recognize the shape and use known formulas to find the area. This is often quicker, especially when the curve forms simple geometric shapes like circles or semicircles.

Semicircle Area

Understanding the area of a semicircle is essential for solving this integral. A semicircle is half of a circle. Thus, its area is half of the area of the full circle. The formula for the area of a circle is \(\pi r^2\). To find the area of a semicircle, we calculate:

• Determine the radius: In our case, it's 2.
• Calculate the full circle area: \(\pi (2)^2 = 4\pi\).
• Divide by 2 for the semicircle: \(\frac{1}{2} \times 4\pi = 2\pi\).

This gives us the area of the semicircle slice that lies between the specified integration limits of \(x = 1\) and \(x = 5\).

Circle Equation

The given integral relates closely to the equation of a circle. The general form of a circle's equation is \( (x-a)^2 + (y-b)^2 = r^2 \), where \( (a, b) \) is the center, and \( r \) is the radius. In this exercise:

• The circle's equation simplifies to \( (x-3)^2 + y^2 = 4 \).
• This indicates:
  • Center at \((3,0)\).
  • Radius of 2, since \( r^2 = 4 \).

The function \(y = \sqrt{4 - (x-3)^2}\) describes the upper half, or semicircle, since this is the positive portion of the circle's equation for \(y\). Recognizing this form allows us to apply geometric methods to calculate the area.

Calculus

Calculus is the mathematical study of change, and it provides the necessary tools to calculate areas under curves using integrals. In integral calculus, when you're calculating a definite integral like \(\int_{1}^{5} \sqrt{4-(x-3)^{2}} \, dx\), you're actually summing up infinitesimally small slices of areas formed under the curve between two points. Thanks to calculus:

• We can understand changes and accumulations, such as areas.
• Definite integrals can be geometrically interpreted as areas under a curve between set boundaries.

This integral, specifically, allows us to apply this concept to find the area of the semicircle using its geometric properties, simplifying what could be a complex calculus problem by recognizing the shape involved.