Question

Find all values of \(c\) that satisfy the Mean Value Theorem for Integrals on the given interval. $$ H(z)=\sin z ; \quad[-\pi, \pi] $$

Short answer

The values of \( c \) are \( c = \sin^{-1} \left( \frac{1}{\pi} \right) \) and \( c = \pi - \sin^{-1} \left( \frac{1}{\pi} \right) \).

Step-by-step solution

State the Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that if a function \( f(x) \) is continuous on the interval \([a, b]\), then there exists at least one number \( c \) in \((a, b)\) such that: \[ f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx \]

Check Continuity of the Function

The given function is \( H(z) = \sin z \). Since the sine function is continuous for all real numbers, it is continuous on the interval \([-\pi, \pi]\).

Evaluate the Integral of the Function

Compute the integral of \( H(z) \) over the interval \([-\pi, \pi]\). \[\int_{-\pi}^{\pi} \sin z \, dz = \left[-\cos z\right]_{-\pi}^{\pi} = -\cos(\pi) - (-\cos(-\pi)) = 2\]

Solve for \( c \) Using the Mean Value Theorem

According to the theorem, there exists a \( c \) for which: \[\sin c = \frac{1}{\pi - (-\pi)} \int_{-\pi}^{\pi} \sin z \, dz = \frac{1}{2\pi} \times 2 = \frac{1}{\pi}\]Now, solve \( \sin c = \frac{1}{\pi} \) for \( c \) in the interval \((-\pi, \pi)\). Since \( \frac{1}{\pi} \) is approximately 0.3183, find the values by using inverse sine: \[c = \sin^{-1} \left(\frac{1}{\pi} \right)\] The principal value is one solution, and for \( \sin \) function more solutions can be found in the interval.

Find All Values of \( c \) Within the Interval

The sine function is periodic with period \(2\pi\). Additionally, \( \sin(c) \) is positive and symmetric around \(z = 0\). Thus, solutions in \((-\pi, \pi)\) will be:\[ c_1 = \sin^{-1} \left( \frac{1}{\pi} \right)\]And the symmetry property provides another solution:\[ c_2 = \pi - \sin^{-1} \left( \frac{1}{\pi} \right)\]

Key concepts

Continuity of Trigonometric Functions

Trigonometric functions, like the sine and cosine, are renowned for their smooth and uninterrupted nature. This characteristic is referred to as continuity. Continuity implies that there are no sharp jumps or breaks where the function is undefined. It ensures that within any given interval, such as \([-\pi, \pi]\), the function can be expected to behave predictably.

- The sine function, \(\sin z\), is defined for all real numbers, making it continuous everywhere. This constant behavior provides a reliable foundation for applying the Mean Value Theorem for Integrals.
- Since the Mean Value Theorem for Integrals requires the function to be continuous over the interval being considered, the sine function perfectly fits these criteria on the interval \([-\pi, \pi]\).

In understanding trigonometric functions, recognizing their continuity is key. It gives us confidence in analyzing their behavior across given intervals, paving the way for further mathematical exploration.

Definite Integrals

Definite integrals play a central role in calculus, representing the accumulation of quantities, like area under a curve, over a specific interval. When we consider a function such as \( H(z) = \sin z \), its integral over an interval gives us valuable insights.
- To compute the integral of \( \sin z \) from \(-\pi\) to \(+\pi\), we evaluate: \\[ \int_{-\pi}^{\pi} \sin z \, dz = \left[-\cos z\right]_{-\pi}^{\pi} = -\cos(\pi) - (-\cos(-\pi)) = 2 \]\
- The result of this integration tells us about the net area between the sine function and the x-axis over the interval \([-\pi, \pi]\).
- The outcome, 2, facilitates the application of the Mean Value Theorem for Integrals, which asserts that there is some point within the interval where the function's average value is "realized."
Understanding definite integrals is crucial as they offer insights into the behavior of functions over specific intervals. This understanding is especially important when utilizing theorems that involve average values.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles when given specific trigonometric values. They serve as the "reverse" of regular trigonometric functions. For the problem at hand, the value \(c\) is found through the inverse of sine.
- The sine inverse, or \(\sin^{-1} \), helps find the angle whose sine is a given number.- For the statement \( \sin c = \frac{1}{\pi} \), we solve for \(c\) in the interval \((-\pi, \pi)\)\: \\[ c = \sin^{-1} \left(\frac{1}{\pi}\right) \]\
- The inverse sine function provides the principal value, which is one solution within our interval, but due to the oscillating nature of the sine function, alternative values exist.
Finding values using inverse trigonometric functions is essential when solving problems involving such equations. They unlock angles from trigonometric ratios, offering solutions within specified bounds.

Periodicity of Trigonometric Functions

The periodicity of trigonometric functions, like sine and cosine, reflects their ability to repeat values over consistent intervals. For sine, this period is \(2\pi\), meaning every \(2\pi\) units, the function begins a new cycle.
- While solving for \(c\) using the Mean Value Theorem for Integrals, the periodic nature of sine is crucial because the function repeats its behavior over intervals.
- Given the solution process for \( \sin c = \frac{1}{\pi} \) in the interval \((-\pi, \pi)\), two solutions emerge: \ - \( c_1 = \sin^{-1} \left( \frac{1}{\pi} \right) \) - \( c_2 = \pi - \sin^{-1} \left( \frac{1}{\pi} \right) \)
The symmetry and periodicity of sine confirm these solutions, capturing how values reappear across the cycle. This periodic nature is vital in understanding and anticipating the behavior of trigonometric functions over continuous intervals.