Question

Verify that a constant \(c\) can be equivalently expressed as a definite integral: (a) \(c \equiv \int_{0}^{b} \frac{c}{b} d x\) (b) \(c=\int_{0}^{c} 1 d t\)

Short answer

Both (a) and (b) show that constant \(c\) can be expressed as these definite integrals, confirming their equivalence.

Step-by-step solution

Understanding Part (a) Setup

We need to verify that the constant \(c\) is equal to the definite integral \(\int_{0}^{b} \frac{c}{b} \, dx\). To do this, we'll begin by understanding and solving the integral.

Evaluate the Integral in Part (a)

Calculate the definite integral: \[ \int_{0}^{b} \frac{c}{b} \, dx = \frac{c}{b} \int_{0}^{b} 1 \, dx. \]Since the integral of 1 over the interval from 0 to \(b\) is \(x\) evaluated from 0 to \(b\), it equals \(b - 0 = b\). So, \[ \frac{c}{b}(b) = c. \]Thus, the definite integral \(\int_{0}^{b} \frac{c}{b} \, dx\) equals \(c\).

Understanding Part (b) Setup

The question asks us to show that \(c = \int_{0}^{c} 1 \, dt\). First, understand that integral as the area under the curve of the constant function \(f(t)=1\) from 0 to \(c\).

Evaluate the Integral in Part (b)

Calculate the definite integral: \[ \int_{0}^{c} 1 \, dt = x \bigg|_{0}^{c}. \]Evaluating this from 0 to \(c\), we have:\[ x \bigg|_{0}^{c} = c - 0 = c. \]Thus, the definite integral \(\int_{0}^{c} 1 \, dt\) equals \(c\).

Conclusion

Both parts (a) and (b) confirm that the constant \(c\) can be represented as these definite integrals. The equivalence is verified through evaluation of each definite integral, showing that both expressions result in the value \(c\).

Key concepts

Constant Function

Let's start with understanding a constant function. A constant function is one where the output value stays the same across all inputs. For example, consider a function defined as \( f(x) = 1 \). This means for every value of \( x \), the function yields an output of 1.

In terms of integrals, when you integrate a constant function over a certain interval, you are essentially measuring the area under a horizontal line. For instance, if you integrate the function \( f(x) = 1 \) from \( x=0 \) to \( x=c \), you create a rectangle with a height of 1 and a width of \( c \). This area, which is the product of the height and width, equals \( c \). Thus, you can express a constant as a definite integral of a constant function, confirming the equivalence \( c = \int_{0}^{c} 1 \ dt \).

Integral Evaluation

Integral evaluation involves finding the exact value of an integral. It involves integrating a function over a specified interval. In the evaluation of definite integrals, you're concerned with calculating the net area under the curve of a function over a set range.

For Part (a), the goal is to evaluate \( \int_{0}^{b} \frac{c}{b} \ dx \). Notice that \( \frac{c}{b} \) is a constant. Integrating it over \( x \) involves multiplying \( \frac{c}{b} \) by the length of the integration interval, \( b \), giving you \( c = \frac{c}{b} \cdot b \). Likewise, for Part (b), evaluating \( \int_{0}^{c} 1 \ dt \) requires you to multiply 1 by the interval length \( c \), directly yielding \( c - 0 = c \).

Both scenarios show a fundamental technique in calculus where finding the value of an integral directly leads to understanding the charge over the interval's position.

Definite Integral Properties

Definite integrals have some interesting properties that simplify solving them. One key property is the ability to integrate a constant. For any constant \( c \), you can compute its integral over the interval \([a, b]\) simply by multiplying \( c \) with the interval length \( (b - a) \). This highlights that integrals can be seen as a way to add up continuous values over an interval.

Another useful property is that if you break down the integral of a function into sums or differences, you can integrate each part separately and then add the results. This property applies when you're dealing with the sum of constant functions or linear combinations thereof.

In the exercises given, these properties allow us to express a constant \( c \) using definite integrals effectively. By applying these properties, the calculations confirm that the integrals simplify directly to the constant \( c \), validating the initial setup of using integrals for such verification.