Problem 22 - Essential Calculus

Question

In Exercises 21-30, sketch the region whose area is given by the definite integral. Then use a geometric formula to evaluate the integral \((a>0, r>0)\) $$ \int_{-a}^{a} 4 d x $$

Step-by-Step Solution

Answer

The region, whose area is given by the definite integral, forms a rectangle of area \(8a\) which equals the value of the integral \(\int_{-a}^{a} 4 dx\).

Step 1: Sketch the region given by the integral

The area given by the integral is between the x-axis and the line \(y=4\) for the interval \([-a, a]\). Being a constant function, the graph is a horizontal straight line at \(y=4\). In the interval \([-a, a]\), this forms a rectangle with length \(2a\) and height \(4\).

Step 2: Evaluate the integral using geometric formula

For a rectangle, the area is calculated as the product of its length and breadth. In this case, length is \(2a\) and breadth is \(4\). Hence the area is \(2a*4=8a\).

Step 3: Calculate the integral

The integral \(\int_{-a}^{a} 4 dx\) is the area under the curve \(f(x) = 4\), we have just calculated to be \(8a\).

Key Concepts

Area Under a Curve

Understanding the area under a curve is essential when studying calculus and mathematical analysis. This concept refers to the region enclosed between a given function and the x-axis over a specific interval. Calculating this area can be visualized as measuring the space under the path of the curve, which is particularly useful in physical contexts, such as determining the distance traveled over time given a speed function.

For the definite integral \(\int_{-a}^{a} 4 dx\), the area under the curve corresponds to a simple geometric shape, since the function represented is a constant. The region is a rectangle extending from \(x = -a\) to \(x = a\) along the x-axis, and from \(y = 0\) to \(y = 4\) on the y-axis. By recognizing this shape, we can bypass complex calculus methods and instead use basic geometry to find the area.

Geometric Formulas

Geometric formulas are mathematical expressions that provide a means to calculate various properties of geometric shapes, like area, volume, perimeter, and length. These formulas are fundamental tools in mathematics as they allow for the simplification of complex problems into manageable solutions.

In our exercise, the area of a rectangle, one of the most elementary geometric shapes, is given by the product of its length and height. The geometric formula used is \(\text{Area} = \text{length} \times \text{height}\). Applying this to the rectangle formed by the definite integral, with a length of \(2a\) and a height of \(4\), gives the area as \(8a\), providing a straightforward approach to solving the integral without performing any actual integration.

Integral Evaluation

The evaluation of an integral is a core concept in calculus, particularly when dealing with the definite integral. While the indefinite integral represents a family of functions, the definite integral provides exact numerical values, thus quantifying the area under a curve between two bounds.

In the given problem, the process of integral evaluation is greatly simplified due to the constant nature of the function. Since \(f(x) = 4\) does not vary with \(x\), evaluating the definite integral \(\int_{-a}^{a} 4 dx\) becomes a matter of multiplying the function's constant value by the width of the interval over which it's integrated. The result, \(8a\), is a precise calculation of the bounded area beneath the horizontal line \(y=4\).

Constant Function

A constant function is a function that always returns the same value, regardless of the input. Mathematically, it is expressed as \(f(x) = C\), where \(C\) is a constant. In the context of calculus, the graph of a constant function is a horizontal line, which simplifies many problems, including the calculation of areas and integrals.

The exercise in question deals with the constant function \(f(x) = 4\), which, when graphed, represents a horizontal line at the y-value of four. Across the interval \( [-a, a]\), this creates a constant height, and applying the integral, we are essentially summing up an infinite number of infinitesimally thin rectangles (or 'slices') of constant height, leading to the simple multiplication of that height \(\text{(4)}\) by the interval width \(\text{(2a)}\) to find the total area. This attribute makes constant functions particularly straightforward when it comes to calculations involving integrals.