Problem 21 - Calculus

Question

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

Step-by-Step Solution

Answer

The values of \( c \) are \( 0, \pi, \) and \( -\pi \).

Step 1: Understand the Mean Value Theorem for Integrals

The Mean Value Theorem (MVT) 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 the interval \([a, b]\) such that: \[ f(c) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \]

Step 2: Apply the Function to the Theorem

In this problem, the function \( H(z) = \sin z \) is given. We will apply the MVT for Integrals with \( a = -\pi \) and \( b = \pi \).

Step 3: Calculate the Integral

Calculate \( \int_{-\pi}^{\pi} \sin z \, dz \). Since \( \int \sin z \, dz = -\cos z \), we have: \[ \int_{-\pi}^{\pi} \sin z \, dz = [-\cos z]_{-\pi}^{\pi} = -\cos(\pi) - (-\cos(-\pi)) = 0. \]

Step 4: Solve for the Mean Value

Use the result from the previous step to find \( f(c) \): \[ f(c) = \frac{1}{\pi - (-\pi)} \cdot 0 = \frac{0}{2\pi} = 0, \] hence \( \sin c = 0 \).

Step 5: Find Values of c

The equation \( \sin c = 0 \) has solutions where \( c = n\pi \) for any integer \( n \). Since \( c \) must be within \([-\pi, \pi]\), the suitable values are \( c = 0, c = \pi, \text{and} \; c = -\pi \).

Key Concepts

Continuous Function

A continuous function is a key concept in understanding many areas of calculus, including the Mean Value Theorem for Integrals. In simple terms, a continuous function is one where you can draw its graph without lifting your pen.
This idea of smoothness is crucial because it ensures there are no sudden jumps or breaks in the function within the given interval.
Mathematical continuity of a function on an interval

• is defined by the property that the limit of the function as it approaches a point is equal to the function's value at that point.
• For example, if a function is continuous on \([a, b]\), then for any point \(c\) in \([a, b]\), we have \(\lim\limits_{x \to c} f(x) = f(c)\).

In our context with the function \(H(z) = \sin z\), this function is continuous across the interval \([-\pi, \pi]\). Because of this continuity, the Mean Value Theorem for Integrals can be applied, reassuring us that there is a certain value or values within the interval where the average value of the function matches the function itself.

Integral Calculus

Integral calculus is a branch of mathematics concerned with the determination of integrals and their properties. When focusing on the Mean Value Theorem for integrals, integral calculus allows us to calculate the area under the curve of a function on a given interval.

The integral of a continuous function from \(a\) to \(b\) gives the net area between the function and the x-axis from point \(a\) to point \(b\).
To find this, we perform a procedure known as integration.

• For our function \(H(z) = \sin z\), the integration \(\int_{-\pi}^{\pi} \sin z \, dz\) results in \((-\cos z)\) evaluated from \(-\pi\) to \(\pi\), yielding a final result of 0.
• This evaluation suggests that the areas above and below the x-axis exactly cancel out over the interval.

Utilizing the result of integration, the theorem confirms that there's at least one value \(c\) in the interval \([-\pi, \pi]\) for which \(f(c)\) equals the average value of the function over that interval.

Trigonometric Functions

Trigonometric functions have fascinating characteristics that make them vital in many calculus problems. These functions, like sine and cosine, are periodic, which means they repeat their values at regular intervals.
For \(H(z) = \sin z\), it completes a full cycle of its wave every \(2\pi\) units.

• One critical property of the function \(\sin z\) is periodicity with zeros at multiples of \(\pi\).
• This is crucial in our problem since, after applying the Mean Value Theorem for Integrals, we found that \(\sin(c) = 0\).

The equation \(\sin c = 0\) typically results in solutions where \(c\) equals \(n\pi\), with \(n\) being an integer. For the interval \([-\pi, \pi]\), the specific solutions are \(c = 0\), \(c = \pi\), and \(c = -\pi\). Understanding this zero crossing trait of \(\sin z\) helps effectively apply and verify the values that satisfy the theorem within the given range.