Question

Evaluate the given integral. $$ \int_{0}^{2 \pi}|\sin 2 x| d x $$

Short answer

The integral evaluates to 4.

Step-by-step solution

Understand the Integral Setup

We need to evaluate the integral \( \int_{0}^{2 \pi} |\sin 2x| \, dx \). The absolute value around \( \sin 2x \) indicates that the integral is divided into sections where \( \sin 2x \) is non-negative and sections where \( \sin 2x \) is negative.

Analyze the Sine Function

The function \( \sin 2x \) completes one full period every \( \pi \) interval. Within the range \([0, 2\pi]\), \( \sin 2x \) completes two full cycles. Determine where \( \sin 2x = 0 \) within this interval by solving \( 2x = n\pi \), where \( n \) is an integer.

Identify Intervals of Positivity and Negativity

The roots are found at \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \). In \([0, \frac{\pi}{2}]\) and \([\pi, \frac{3\pi}{2}]\), \( \sin 2x \geq 0 \) and in \([\frac{\pi}{2}, \pi] \) and \([\frac{3\pi}{2}, 2\pi] \), \( \sin 2x < 0 \). We should adjust the function to \( -\sin 2x \) in these intervals.

Set Up the Integral

To handle the absolute value, break the integral into four parts:- \( \int_{0}^{\frac{\pi}{2}} \sin 2x \, dx \)- \( \int_{\frac{\pi}{2}}^{\pi} -\sin 2x \, dx \)- \( \int_{\pi}^{\frac{3\pi}{2}} \sin 2x \, dx \)- \( \int_{\frac{3\pi}{2}}^{2\pi} -\sin 2x \, dx \)

Integrate Each Part

The integral of \( \sin 2x \) is \(-\frac{1}{2}\cos 2x\). Calculate each partition:1. \( \int_{0}^{\frac{\pi}{2}} \sin 2x \, dx = [-\frac{1}{2}\cos 2x]_0^{\frac{\pi}{2}} = 1 \)2. \( \int_{\frac{\pi}{2}}^{\pi} -\sin 2x \, dx = \frac{1}{2}[\cos 2x]_{\frac{\pi}{2}}^{\pi} = 1 \)3. \( \int_{\pi}^{\frac{3\pi}{2}} \sin 2x \, dx = [-\frac{1}{2}\cos 2x]_{\pi}^{\frac{3\pi}{2}} = 1 \)4. \( \int_{\frac{3\pi}{2}}^{2\pi} -\sin 2x \, dx = \frac{1}{2}[\cos 2x]_{\frac{3\pi}{2}}^{2\pi} = 1 \)

Sum the Integral Results

Add the results of each of the four integrals:\( 1 + 1 + 1 + 1 = 4 \). Thus, the value of the integral is 4.

Key concepts

Definite Integrals

A definite integral is a fundamental concept in calculus representing the area under a curve within a specific interval on the x-axis. This is uniquely valuable because, unlike indefinite integrals which have no set limits, definite integrals have specified bounds, allowing for precise calculations. The process involves calculating the net area, taking into consideration where the function lies above or below the x-axis.

For example, when we evaluate \( \int_{a}^{b} f(x) \, dx \), we are interested in the integral of \( f(x) \) from \( x = a \) to \( x = b \). This is computed using the fundamental theorem of calculus, which relates differentiation and integration, showing that such an integral can be derived from an anti-derivative \( F(x) \) of \( f(x) \).

In our original problem, the definite integral is from \( 0 \) to \( 2\pi \), and it considers the function \( |\sin 2x| \), thus requiring the calculation of area for each interval."},{

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are periodic functions that repeat their values in regular intervals, which makes them crucial in understanding cyclical phenomena. These functions are defined based on the angles of right triangles and are periodic with periods of \( 2\pi \) for sine and cosine. For the function \( \sin 2x \), its period is reduced to \( \pi \) since multiplying the angle by 2 causes the sine wave to complete a cycle in half the time.

These functions have certain symmetries and properties that are essential when calculating integrals involving trigonometric terms. For instance, within each cycle of \( \sin 2x \):

• The function crosses zero at specific points: \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \) and \( 2\pi \).
• It exhibits positive values over some intervals (such as \([0, \frac{\pi}{2}])\) and negative values over others (such as \([\frac{\pi}{2}, \pi])\).

Understanding these properties helps in setting up the integral correctly, ensuring accurate calculations when areas need to be adjusted for negative values.

Absolute Value in Integrals

Dealing with integrals that involve absolute values can be tricky. The key is to recognize that an absolute value alters the function to ensure all outputs are non-negative. When you integrate an absolute value function, you will often need to break the integral into segments where the inner function is positive and where it is negative.

For the integral \( \int_{0}^{2 \pi} |\sin 2x| \, dx \), notice how \( \sin 2x \) alternates between positive and negative values periodically. This requires splitting the integral into parts where:

• \( \sin 2x \) is non-negative: leave the function as is.
• \( \sin 2x \) is negative: replace it with \(-\sin 2x\) to adjust for this.

In the problem, it was split into four segments, adjusting where \( \sin 2x \) is negative, and each part is integrated separately. The results are summed to get the total area under the curve, thus providing an accurate valuation of the original integral.