91Ó°ÊÓ

A constant function is a simple yet important concept in calculus and algebra. It is defined as a function that returns the same value for any input within its domain. This means that if you have a function defined as \( f(x) = c \), where \( c \) is a constant, then no matter what value of \( x \) you substitute, \( f(x) \) will always equal \( c \).
In this exercise, we are dealing with the constant function \( f(x) = 6 \). Graphically, a constant function is represented by a horizontal line on the Cartesian plane. This line is parallel to the x-axis, and its intersection with the y-axis is at the y-value equal to the constant, which in this case is 6. The simplicity of a constant function makes it particularly straightforward when evaluating a definite integral, as the function's value does not vary over the interval you are examining.

The interval length is a measure of the distance between two points on the x-axis, specified as the bounds of integration. When calculating an integral, understanding the interval over which you're integrating is crucial.
In the given problem, the integral \( \int_{-2}^{5} 6 \, dx \) spans the interval from \( x = -2 \) to \( x = 5 \). To find the length of this interval, you subtract the starting point from the endpoint:

• Formula: \( L = b - a \) where \( a \) and \( b \) are the lower and upper limits of the interval, respectively.

Plugging in the values, we have:

• \( L = 5 - (-2) = 5 + 2 = 7 \)

This interval length tells us that the integration must encompass the distance of 7 units along the x-axis, which directly affects the result of the integral for a constant function.

The area under a curve represents the value obtained from the definite integral of a function between two bounds. More intuitively, if the function represents a real-world context (like speed), the area could represent a real quantity (like distance). For a constant function, this calculation is straightforward because the 'curve' is actually a straight, horizontal line.
In our exercise, the definite integral \( \int_{-2}^{5} 6 \, dx \) evaluates the area under the constant function \( f(x) = 6 \) between the points \( x = -2 \) and \( x = 5 \).

• This area is equivalent to a rectangle where:
• The width is the interval length, which is 7 units.
• The height is the constant value of the function, which is 6 units.

Therefore, the area under the curve, or the area of the rectangle in this context, is calculated by simply multiplying these two dimensions:

• \( \text{Area} = 6 \times 7 = 42 \)

This result demonstrates how the concept of integrating a constant function can be simplified into finding the area of a geometric shape.