Problem 31 - Calculus and its applications

Question

Use geometry to evaluate each definite integral. \(\int_{2}^{6} 3 d x\)

Step-by-Step Solution

Answer

The integral evaluates to 12.

Step 1: Identify the Geometric Shape

The function we are dealing with is a constant function, \( f(x) = 3 \). When we graph this function from \( x = 2 \) to \( x = 6 \), we form a horizontal line at \( y = 3 \). The region under this line from \( x = 2 \) to \( x = 6 \) is a rectangle.

Step 2: Calculate the Dimensions of the Rectangle

The height of the rectangle is given by the constant value of the function, which is 3. The width of the rectangle is the difference between the upper and lower limits of the integral, which is \( 6 - 2 = 4 \).

Step 3: Compute the Area of the Rectangle

The area of a rectangle is given by the formula \( \text{Area} = \text{height} \times \text{width} \). Here, the height is 3, and the width is 4. So, \( \text{Area} = 3 \times 4 = 12 \).

Step 4: Interpret the Area as the Value of the Integral

The definite integral \( \int_{2}^{6} 3 \, dx \) represents the area under the curve of \( f(x) = 3 \) between \( x = 2 \) and \( x = 6 \). Therefore, the value of the integral is 12.

Key Concepts

Geometry in Calculus

Geometry and calculus come together beautifully to help us understand the areas and shapes that are formed when we analyze functions. In calculus, when we talk about integrating a function, what we're often doing is finding the area under a curve represented by that function.
This idea of using geometric shapes to solve integrals is called geometric integration. It's especially useful in cases where the function leads to simple geometric regions such as rectangles, triangles, or circles. These shapes have well-known area formulas, which can make solving the integral straightforward.
In our example, the function is constant, which forms a simple rectangle on the coordinate plane. Recognizing this link to a rectangle allows us to quickly determine the definite integral by using basic geometry to calculate the area.

Area Under a Curve

Finding the area under a curve is a key part of understanding integrals in calculus. A definite integral gives you precisely this: the area between a function's curve, the x-axis, and the limits of integration.
The area is calculated differently depending on the shape made by the curve. However, the basic principle remains the same — you're essentially measuring how much space is occupied beneath the curve. For different types of functions, this area can look different:

• For linear functions, it might form a trapezoid.
• For quadratic functions, it might form a parabolic section.
• For constant functions, like in our exercise, it forms a simple rectangle.

All these are just different ways to visually understand what the integral measures, which is the accumulated value of the function over a certain interval.

Constant Functions

A constant function is one that stays the same no matter what the input is. This means it plots as a horizontal line on a graph. For example, the function in our exercise, \( f(x) = 3 \), remains at 3 for any value of \( x \).
When integrating a constant function over a specific interval, the process is made easier because the shape we need to consider is always a rectangle. This is because the function doesn't change; it consistently produces the same output, resulting in a flat horizontal line.
The area under this horizontal line (the definite integral) is simply the height of this line times the width of the interval. In the example given:

• The height (the value of the function) is 3.
• The width (the interval from 2 to 6) is 4.

Therefore, the integral's value is just \(3 \times 4 = 12\), illustrating how easily areas can be calculated using constant functions.