Problem 30 - Calculus and its applications

Question

Use geometry to evaluate each definite integral. \(\int_{0}^{5} 6 d x\)

Step-by-Step Solution

Answer

The value of the integral is 30.

Step 1: Understanding the Integral as an Area

The integral \( \int_{0}^{5} 6 \, dx \) represents the area under the curve \( y = 6 \) from \( x = 0 \) to \( x = 5 \). Since this is a constant function, the graph is a horizontal line.

Step 2: Visualize the Shape

Visualize the area under the line \( y = 6 \) from \( x = 0 \) to \( x = 5 \). This forms a rectangle with a base along the \( x \)-axis from 0 to 5 and a height of 6.

Step 3: Calculate the Area of the Rectangle

To find the area of a rectangle, use the formula \( \text{Area} = \text{base} \times \text{height} \). Here, the base is 5 (from 0 to 5) and the height is 6, so the area is \( 5 \times 6 = 30 \).

Step 4: Interpret the Result

The definite integral \( \int_{0}^{5} 6 \, dx = 30 \) confirms that the area under the curve from \( x = 0 \) to \( x = 5 \) is 30.

Key Concepts

Geometry in Calculus

In calculus, geometry plays a pivotal role, especially when dealing with definite integrals. A definite integral, such as \( \int_{0}^{5} 6 \, dx \), is fundamentally concerned with computing the total area under a curve from one point to another. Unlike indefinite integrals, which are more abstract, definite integrals have a direct geometric meaning. Here, they tell us the exact area bounded by the curve, the \( x \)-axis, and the vertical lines at the bounds of integration.
By thinking of a definite integral as a geometric figure, you can translate complex calculus problems into simpler geometric shapes. This method is key because it not only helps in visualizing the problem but also in providing a tangible solution without solely relying on algebraic manipulations. The geometry in calculus is essentially our bridge to convert abstract expressions into understandable and often rectangular shapes or areas.

Area Under a Curve

The concept of the area under a curve is crucial in integrating functions. When dealing with functions like \( y = 6 \), where the line is horizontal and does not change, the area calculation becomes straightforward. The area under the curve, or more appropriately the line in this case, from \( x = 0 \) to \( x = 5 \), can simply be visualized as a large rectangle.
For more complex curves, you might imagine this area as comprised of many tiny rectangles, helping in the calculation process. In cases like the definite integral \( \int_{0}^{5} 6 \, dx \), the problem simplifies further because the 'curve' is actually a flat line. The goal is always the same: sum up the area occupied between the curve and the \( x \)-axis over an interval to find the integral.

• The base of this shape is the difference in \( x \)-values: \( 5 - 0 = 5 \).
• The height is the constant value of the function, 6.
• Multiplying these gives the area: \( 5 \times 6 = 30 \).

Rectangular Regions

Rectangular regions are one of the most straightforward areas to understand when computing definite integrals. Specifically, when the integrand is a constant, the region under the curve forms a perfect rectangle.
This is easily identifiable in an integral like \( \int_{0}^{5} 6 \, dx \), where the base is the interval \( [0, 5] \) and the constant function value \( y = 6 \) determines the height of the rectangle. To find the total area—akin to the value of the definite integral—you multiply these dimensions just as you would in basic geometry.
The use of rectangular regions demystifies many integral problems, transforming them into simple arithmetic once you understand the shape that the integral represents. This approach is particularly helpful for beginners to comfortably bridge their understanding of basic geometrical concepts with more abstract calculus principles.