Problem 104 - Calculus and its applications

Question

Evaluate. $$ \int_{-2}^{2} \sqrt{4-x^{2}} d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Integral Type

The given integral \( \int_{-2}^{2} \sqrt{4-x^2} \, dx \) resembles the formula for the area of a semicircle. Recognize that \( \sqrt{4-x^2} \) represents the upper half of a circle with radius 2, centered at the origin.

Step 2: Set up the Circle Equation

The general formula for a circle is \( x^2 + y^2 = r^2 \). Here, \( r = 2 \), so the equation of the circle is \( x^2 + y^2 = 4 \). Solving for \( y \), we find \( y = \sqrt{4 - x^2} \), which represents the top semicircle.

Step 3: Determine Bounds of Integration

The limits of the integral are from \(-2\) to \(2\), which are the points where the circle intersects the x-axis. These are also the points where the semicircle spans horizontally.

Step 4: Calculate the Area Under the Semicircle

The integral represents the area of the upper semicircle. The formula for the area of a semicircle is \( \frac{1}{2} \pi r^2 \). Substituting \( r = 2 \), the area is \( \frac{1}{2} \pi (2^2) = 2\pi \).

Step 5: Conclude the Evaluation

Thus, the value of the integral \( \int_{-2}^{2} \sqrt{4-x^2} \, dx \) is the area of the semicircle, which equals \( 2\pi \).

Key Concepts

Definite Integrals

In the world of calculus, definite integrals are a fundamental concept that help us find the area under a curve. Unlike indefinite integrals, which produce a family of functions, definite integrals provide a specific numerical value. They do this by calculating the accumulated area between the curve and the x-axis over a given interval. This means when you work with a definite integral, you integrate from one point to another.
This specific point interval gives us very useful information about particular sections of the function. The integral \( \int_{-2}^{2} \sqrt{4-x^{2}} \, dx \) is a great example. It asks for the calculation from \( x = -2 \) to \( x = 2 \). The boundaries of this integral directly correspond to the limits of an area we're interested in, essentially capturing the entire semicircle's surface in this case.

• The lower limit of integration is \( -2 \).
• The upper limit of integration is \( 2 \).
• The definite integral computes the total area in this range.

Geometric Interpretation

Integral calculus provides a unique way of understanding geometry. It lets us not just look at shapes, but also quantify them in terms of areas. When looking at the integral \( \int_{-2}^{2} \sqrt{4-x^{2}} \, dx \), we must interpret this as a geometric problem.
On the coordinate plane, the function \( y = \sqrt{4-x^{2}} \) graphs the top half of a circle (semicircle) centered at the origin. This semicircle curve provides an excellent geometric interpretation.

• The function's output, \( y = \sqrt{4-x^{2}} \), represents the curve of the semicircle.
• The curve spans between \( x = -2 \) and \( x = 2 \), exactly our limits of integration.
• The geometric shape here is a semicircle with the x-axis as its baseline.

Area of a Circle

Understanding the formula for a circle's area is crucial in integral calculus, especially when tackling tasks like evaluating the given integral. A circle's area formula is \( \pi r^2 \), where \( r \) represents the radius of the circle. To solve the integral \( \int_{-2}^{2} \sqrt{4-x^{2}} \, dx \), we recognize that it represents the area of a semicircle.
Let's break this down:

• The equation \( x^2 + y^2 = r^2 \) stands for a full circle.
• For this exercise, assume \( r = 2 \), thus our circle's equation is \( x^2 + y^2 = 4 \).
• The function \( y = \sqrt{4-x^{2}} \) represents simply the upper half, or semicircle.
• The area of a semicircle is half the area of a whole circle, calculated as \frac{1}{2} \pi r^2\.
• Substitute \( r = 2 \) in, giving us an area of \( 2\pi \).

These calculations lead us to conclude that the integral value, \( \int_{-2}^{2} \sqrt{4-x^{2}} \, dx = 2\pi \), which is the area of this semicircle.