Definition of the Absolute Value Function

Oct 13, 2025 by vinooalluri.com 138 views

Introduction

In this article, we will evaluate the definite integral of the absolute value function $|x-1|$ from $0$ to $2$ . The absolute value function is a fundamental concept in mathematics, and its integral is a crucial tool in calculus. We will use the definition of the absolute value function and the properties of definite integrals to evaluate this integral.

Definition of the Absolute Value Function

The absolute value function is defined as:

|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

In this case, we have $|x-1|$ , which can be rewritten as:

|x-1| = \begin{cases} x-1 & \text{if } x-1 \geq 0 \\ -(x-1) & \text{if } x-1 < 0 \end{cases}

Evaluating the Integral

To evaluate the integral, we need to consider two cases: when $x-1 \geq 0$ and when $x-1 < 0$ .

Case 1: $x-1 \geq 0$

In this case, we have $x \geq 1$ , and the absolute value function can be rewritten as:

|x-1| = x-1

The integral becomes:

\int_0^2|x-1| \, dx = \int_0^2 (x-1) \, dx

Using the power rule of integration, we get:

\int_0^2 (x-1) \, dx = \left[\frac{x^2}{2} - x\right]_0^2

Evaluating the expression at the limits of integration, we get:

\left[\frac{x^2}{2} - x\right]_0^2 = \left(\frac{2^2}{2} - 2\right) - \left(\frac{0^2}{2} - 0\right)

Simplifying the expression, we get:

\left(\frac{2^2}{2} - 2\right) - \left(\frac{0^2}{2} - 0\right) = 2 - 2 = 0

Case 2: $x-1 < 0$

In this case, we have $x < 1$ , and the absolute value function can be rewritten as:

|x-1| = -(x-1)

The integral becomes:

\int_0^2|x-1| \, dx = \int_0^2 -(x-1) \, dx

Using the power rule of integration, we get:

\int_0^2 -(x-1) \, dx = -\left[\frac{x^2}{2} - x\right]_0^2

Evaluating the expression at the limits of integration, we get:

-\left[\frac{x^2}{2} - x\right]_0^2 = -\left(\frac{2^2}{2} - 2\right) + \left(\frac{0^2}{2} - 0\right)

Simplifying the expression, we get:

-\left(\frac{2^2}{2} - 2\right) + \left(\frac{0^2}{2} - 0\right) = -2 + 2 = 0

Conclusion

In both cases, we found that the integral of the absolute value function $|x-1|$ from $0$ to $2$ is equal to $0$ . This result makes sense, since the absolute value function is symmetric around the point $x=1$ , and the integral is taken over a symmetric interval.

Final Answer

The final answer is $\boxed{0}$ .

Discussion

The absolute value function is a fundamental concept in mathematics, and its integral is a crucial tool in calculus. In this article, we evaluated the definite integral of the absolute value function $|x-1|$ from $0$ to $2$ . We used the definition of the absolute value function and the properties of definite integrals to evaluate this integral. We found that the integral is equal to $0$ , which makes sense, since the absolute value function is symmetric around the point $x=1$ , and the integral is taken over a symmetric interval.

References

[1] "Calculus" by Michael Spivak
[2] "Introduction to Calculus" by Michael Artin
[3] "Mathematics for the Nonmathematician" by Morris Kline

Introduction

In our previous article, we evaluated the definite integral of the absolute value function $|x-1|$ from $0$ to $2$ . In this article, we will answer some common questions related to this topic.

Q: What is the absolute value function?

A: The absolute value function is a mathematical function that returns the distance of a number from zero on the number line. It is defined as:

|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

Q: How do you evaluate the definite integral of the absolute value function?

A: To evaluate the definite integral of the absolute value function, you need to consider two cases: when $x-1 \geq 0$ and when $x-1 < 0$ . In the first case, the absolute value function can be rewritten as $x-1$ , and the integral becomes:

\int_0^2|x-1| \, dx = \int_0^2 (x-1) \, dx

In the second case, the absolute value function can be rewritten as $-(x-1)$ , and the integral becomes:

\int_0^2|x-1| \, dx = \int_0^2 -(x-1) \, dx

Q: Why do we need to consider two cases?

A: We need to consider two cases because the absolute value function changes its behavior at the point $x=1$ . When $x \geq 1$ , the absolute value function is equal to $x-1$ , and when $x < 1$ , the absolute value function is equal to $-(x-1)$ .

Q: What is the final answer to the definite integral of the absolute value function?

A: The final answer to the definite integral of the absolute value function is $\boxed{0}$ .

Q: Why is the final answer equal to 0?

A: The final answer is equal to 0 because the absolute value function is symmetric around the point $x=1$ , and the integral is taken over a symmetric interval.

Q: What are some common applications of the absolute value function?

A: The absolute value function has many applications in mathematics, physics, and engineering. Some common applications include:

Modeling the distance between two points on a number line
Representing the magnitude of a vector
Describing the behavior of a physical system that has a minimum or maximum value

Q: What are some common mistakes to avoid when evaluating the definite integral of the absolute value function?

A: Some common mistakes to avoid when evaluating the definite integral of the absolute value function include:

Failing to consider the two cases
Not rewriting the absolute value function correctly
Not evaluating the integral at the correct limits of integration

Conclusion

In this article, we answered some common questions related to the definite integral of the absolute value function. We hope that this article has been helpful in clarifying any confusion and providing a better understanding of this important mathematical concept.

Final Answer

The final answer is $\boxed{0}$ .

References

[1] "Calculus" by Michael Spivak
[2] "Introduction to Calculus" by Michael Artin
[3] "Mathematics for the Nonmathematician" by Morris Kline