Problem 37 - Calculus: Early Transcendentals

Question

Evaluate the integral by interpreting it in terms of areas. \( \displaystyle \int^0_{-3} \bigl( 1 + \sqrt{9 - x^2} \bigr) \, dx \)

Step-by-Step Solution

Answer

The integral evaluates to \(3 + \frac{9\pi}{2}\).

Step 1: Understand the Components of the Integral

The integral \( \int^0_{-3} \bigl( 1 + \sqrt{9 - x^2} \bigr) \, dx \) consists of two parts: 1 and \( \sqrt{9 - x^2} \). The term '1' represents a constant function, while \( \sqrt{9 - x^2} \) represents a semicircle with radius 3 centered at the origin, covering only the upper half in the x-y plane.

Step 2: Analyze the Constant Function

The constant function '1' over the interval from \(-3\) to 0 is a horizontal line. The area under this line from \(-3\) to 0 is a rectangle with a width of 3 and a height of 1. Therefore, the area is \(3 \times 1 = 3\).

Step 3: Area of the Semicircle

The expression \( \sqrt{9 - x^2} \) describes the top half of a circle with a radius of 3. The entire area of a circle with radius 3 is \( \pi \times 3^2 = 9\pi \). Since we only have the top half (semicircle) from \(-3\) to 0, the area is half of the full circle: \( \frac{1}{2} \times 9\pi = \frac{9\pi}{2} \).

Step 4: Combine the Areas

Summing the areas calculated, we get the total area under the curve \(1 + \sqrt{9 - x^2} \) from \(-3\) to 0. Therefore, the total area (integral value) is \(3 + \frac{9\pi}{2}\).

Key Concepts

Definite Integrals

Definite integrals are a fundamental part of integral calculus and help in determining the area under a curve within a specific interval. Unlike indefinite integrals, which yield a family of functions, definite integrals provide a specific numerical value. This value represents the net area between the curve and the x-axis over the given interval.
The definite integral is denoted by the integral symbol with upper and lower limits: \( \int_{a}^{b} f(x) \, dx \). Here, \(a\) and \(b\) are the limits of integration, signifying the interval over which the area is calculated.

• If the function is above the x-axis, the integral sums up the area as positive.
• If below, the area counts as negative towards the net result.

For example, in the exercise provided, we compute the integral from -3 to 0 of the function \(1 + \sqrt{9 - x^2}\). This integral determines the total area consisting of the part under the semicircle and the flat area represented by the constant 1 between these bounds.

Geometric Interpretation

The geometric interpretation is a vital concept for understanding integrals more intuitively. It involves visualizing the definite integral as the accumulation of infinitesimally small areas under a curve over an interval. This visualization translates mathematical expressions into shapes and areas you can easily understand.
In our exercise, the function \(1 + \sqrt{9 - x^2}\) consists of two parts with distinct geometric meanings:

• The '1' represents a horizontal line from \(-3\) to 0, forming a rectangle area of 3 square units.
• The expression \(\sqrt{9 - x^2}\) outlines the top half of a circle, known as a semicircle, with a radius of 3. It spans over the same interval.

Combining these, you sum the area of the rectangle and half the circle to get a full understanding of the total area captured by the definite integral between the limits in question.

Area Under a Curve

Determining the "area under a curve" is a classic application of definite integrals. This concept tells us how much "space" is under a curve on a graph within a certain interval on the x-axis. This can represent various physical quantities like distance or probability, depending on the context.
In our example, the function \(1 + \sqrt{9 - x^2} \) includes two main components:

• The constant function '1', which creates a rectangular region with an easy-to-calculate area: width 3 (from \(-3\) to 0) times height 1 equals 3 square units.
• A semicircle described by \(\sqrt{9 - x^2} \). By recognizing this as the top half of a circle with radius 3, you can easily determine its area as half of \(9\pi\) (the area of the whole circle), yielding \(\frac{9\pi}{2}\).

Thus, by summing the outcomes of these two parts, we find the total area under the curve from \(-3\) to 0, perfectly illustrating the use of definite integrals to find areas.