Problem 28 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int_{0}^{\pi} \frac{1}{2}(\cos x+|\cos x|) d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \).

Step 1: Understanding the Integral

The given integral is \( \int_{0}^{\pi} \frac{1}{2} (\cos x + |\cos x|) \, dx \). We need to evaluate this integral over the interval \([0, \pi]\). Notice that \( |\cos x| \) affects the expression of the function inside the integral, so we will consider the behavior of \( \cos x \) over the interval \([0, \pi]\).

Step 2: Identify the Interval Behavior

On the interval \([0, \pi/2]\), \( \cos x \) is non-negative, so \(|\cos x| = \cos x\). Hence the integral becomes \( \int_{0}^{\pi/2} \cos x \, dx \). On the interval \([\pi/2, \pi]\), \( \cos x \) is non-positive, so \(|\cos x| = -\cos x\). Thus, in this interval, the integral is \( \int_{\pi/2}^{\pi} 0 \, dx \), as the expression \( \cos x + | \cos x| \) becomes zero.

Step 3: Calculate the First Interval Integral

Evaluate \( \int_{0}^{\pi/2} \cos x \, dx \). The antiderivative of \( \cos x \) is \( \sin x \). Thus, \[ \int_{0}^{\pi/2} \cos x \, dx = \left[ \sin x \right]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1. \]

Step 4: Calculate the Second Interval Integral

For the interval \([\pi/2, \pi]\), the integral value is simply zero because the integrand function becomes zero in this case: \( \int_{\pi/2}^{\pi} 0 \, dx = 0 \).

Step 5: Combine the Results

Add the results of both intervals together. The integral \( \int_{0}^{\pi} \frac{1}{2} (\cos x + | \cos x |) \, dx \) simplifies to \( \frac{1}{2} \times (1 + 0) = \frac{1}{2} \). Thus, the evaluated integral is \( \frac{1}{2} \).

Key Concepts

Definite Integrals

When working with integrals, you are often dealing with the concept of definite integrals. A definite integral provides the area under a curve between two specified points on a given function. In our exercise, the definite integral is evaluated from 0 to \( \pi \). This area calculation becomes essential in understanding how functions behave over specific intervals.

• The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper bounds, respectively.
• The process involves finding an antiderivative of the integrand \( f(x) \), then computing the difference \( F(b) - F(a) \).

In evaluating definite integrals, consider the behavior of the function within the given limits. Here, the integral involves the function \( \frac{1}{2}(\cos x + |\cos x|) \), which requires careful examination of the properties of \( \cos x \), especially due to the presence of the absolute value.

Trigonometric Functions

Trigonometric functions, such as \( \cos x \), play a pivotal role in calculus for their unique properties and periodic behavior. The cosine function, \( \cos x \), oscillates between -1 and 1, providing mathematical models for various natural phenomena. Understanding trigonometric functions is essential in calculating integrals involving them. Let's delve into the behavior of \( \cos x \) in our integral:

• On the interval \([0, \pi/2]\), the function \( \cos x \) remains positive and decreases from 1 to 0.
• In contrast, on \([\pi/2, \pi]\), \( \cos x \) is negative, transitioning from 0 to -1.

It's crucial to analyze these intervals to determine how absolute values affect the function. Through this analysis, we segment the integral appropriately, making complex expressions simpler to resolve by hand or computationally.

Absolute Value Functions

The absolute value function modifies the behavior of trigonometric functions within integrals. Here, specifically for the exercise, \(|\cos x|\) influences the integral's solution. Absolute value represents the distance of a number from zero on the number line, regardless of direction (positive/negative).- For expressions with \(|\cos x|\), the function \(|\cos x| = cos x\) when \(\cos x\) is positive.- Conversely, \(|\cos x| = -\cos x\) when \(\cos x\) is negative.

This dual nature results in a piecewise evaluation, separating the integral into parts where the function behaves differently. Accurately determining these segments ensures the correct evaluation of the integral. By understanding which portions of the interval result in non-negative vs. non-positive outcomes, we can simplify such integrals even with the presence of absolute values impacting the function.