State whether this is true or false. If true, give an explanation. If false, give a counterexample. ### Improper Integrals: Convergence Example#### (b) The improper integral $\int_{-\infty}^{\infty} \sin x \, dx$ converges and is equal to 0.This statement illustrates an example of an improper integral. Specifically, it examines the integral of the sine function over the entire real line, from negative infinity to positive infinity. Despite the oscillatory nature of the sine function, the integral converges to zero.The improper integral we are considering is:\[ \int_{-\infty}^{\infty} \sin x \, dx \]### Key Concepts:1. **Improper Integral**: An integral with one or both limits extending to infinity, or an integral over an interval where the integrand becomes unbounded.2. **Convergence**: The concept that an improper integral has a finite value, even though it extends over an infinite domain or involves an unbounded function.### Explanation:The sine function oscillates between -1 and 1 perpetually. When integrating $\sin x$ over a symmetric range around the origin, the positive and negative contributions cancel out due to the symmetry. This balance results in the integral converging to zero.### In-Depth Study:For a deeper understanding, you can break this integral down as follows:\[ \int_{-\infty}^{\infty} \sin x \, dx = \lim_{A \to \infty} \left( \int_{-A}^{A} \sin x \, dx \right)\]Using the antiderivative of $\sin x$, which is $-\cos x$, evaluate the integral:\[ \int_{-A}^{A} \sin x \, dx = \left[ -\cos x \right]_{-A}^{A} = -\cos A - (-\cos (-A)) = -\cos A - (-\cos A) = -\cos A + \cos A = 0\]Thus, the integral of $\sin x$ over an infinite domain indeed converges to zero.

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Answered: (b) The improper integral sin x dx…

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(b) The improper integral sin x dx converges and is equal to 0.

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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State whether this is true or false. If true, give an explanation. If false, give a counterexample.

### Improper Integrals: Convergence Example

#### (b) The improper integral $\int_{-\infty}^{\infty} \sin x \, dx$ converges and is equal to 0.

This statement illustrates an example of an improper integral. Specifically, it examines the integral of the sine function over the entire real line, from negative infinity to positive infinity. Despite the oscillatory nature of the sine function, the integral converges to zero.

The improper integral we are considering is:
\[ \int_{-\infty}^{\infty} \sin x \, dx \]

### Key Concepts:
1. **Improper Integral**: An integral with one or both limits extending to infinity, or an integral over an interval where the integrand becomes unbounded.
2. **Convergence**: The concept that an improper integral has a finite value, even though it extends over an infinite domain or involves an unbounded function.

### Explanation:
The sine function oscillates between -1 and 1 perpetually. When integrating $\sin x$ over a symmetric range around the origin, the positive and negative contributions cancel out due to the symmetry. This balance results in the integral converging to zero.

### In-Depth Study:
For a deeper understanding, you can break this integral down as follows:
\[ \int_{-\infty}^{\infty} \sin x \, dx = \lim_{A \to \infty} \left( \int_{-A}^{A} \sin x \, dx \right)\]
Using the antiderivative of $\sin x$, which is $-\cos x$, evaluate the integral:
\[ \int_{-A}^{A} \sin x \, dx = \left[ -\cos x \right]_{-A}^{A} = -\cos A - (-\cos (-A)) = -\cos A - (-\cos A) = -\cos A + \cos A = 0\]

Thus, the integral of $\sin x$ over an infinite domain indeed converges to zero.

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