Problem 21 - 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 \((a>0, r>0)\) $$ \int_{0}^{3} 4 d x $$

Step-by-Step Solution

Answer

The answer is 12.

Step 1: Identifying the Function and the Interval

The integral given is \(\int_{0}^{3} 4 dx\) . This integral represents the area under the constant function \(f(x) = 4 \) from \(x=0\) to \(x=3\). This forms a rectangle with width equal to 3 units and height 4 units.

Step 2: Sketching the Region

On drawing it on a graph, you can see a rectangle with the base on the x-axis ranging from x=0 to x=3, and the height reaching up to y=4 on the y-axis.

Step 3: Calculating the Area using the Geometric Formula

The area \(A\) of a rectangle is given by the formula \(A = length * width\). Applying this, the area under the line \(f(x) = 4\) from \(x=0\) to \(x=3\) is \(A = 4 * 3\).

Key Concepts

Geometric Interpretation of Definite Integrals

When we deal with definite integrals, we are often exploring the concept of the area under a curve on a graph. The geometric interpretation of a definite integral is a fundamental idea in calculus, and it's tied closely to the practical applications of integration. In the case of the exercise problem \( \int_{0}^{3} 4 dx \) we are looking at a very simple example. This integral asks us to consider the area located between the x-axis, the lines x=0, x=3, and the constant function \( f(x) = 4 \).

This is an easy-to-visualize scenario and forms a perfect rectangle on the Cartesian plane. Understanding this geometric aspect is crucial as it forms the basis for interpreting more complex integrals, where the shape under the curve isn't a simple geometric figure. In such situations, definite integrals allow us to calculate areas that are not easily handled by basic geometric formulas.

Therefore, the definite integral essentially provides us with a tool to aggregate infinitely small pieces of area, which combined, give us the total area under a curve between two points on the x-axis. This precise accumulation is the power of the integral calculus, revealing how it extends beyond simple geometrical shapes.

Area Under a Curve

Calculating the area under a curve is a classic application of integrals. The area under a curve is the region enclosed by the graph of a function, the x-axis, and the vertical lines at the endpoints of the interval. This measure of space is highly relevant in disciplines such as physics, economics, and engineering, where it can represent quantities like distance, total revenue, or work done.

In the given exercise, we looked at the definite integral \( \int_{0}^{3} 4 dx \), which represents the area under the horizontal line \( f(x) = 4 \) from \( x = 0 \) to \( x = 3 \). This specific case simplifies to calculating the area of a rectangle since the

Rectangular Area Calculation

In geometry, calculating the area of a rectangle is one of the most straightforward tasks, involving the use of the formula \( A = length \times width \). This process becomes extremely relevant in integral calculus when we deal with the area under a curve that takes the shape of a rectangle.

In the exercise \( \int_{0}^{3} 4 dx \) we have exactly such a case. Here, the rectangle has a width from \( x=0 \) to \( x=3 \) (which is the length of the base on the x-axis) and a height given by the constant function value \( f(x) = 4 \). Applying our rectangular area formula yields \( A = 4 \times 3 = 12 \) square units.

This calculation showcases the simplicity and elegance which definite integrals can offer when applied to calculate areas for basic shapes. It's a neat transition from geometric principles to calculus concepts and sets the foundation for understanding more complex areas where shapes are not regular, and the function describing the curve is not constant.