Problem 109 - Calculus Volume 1

Question

In the following exercises, use the comparison theorem. Show that \(\int_{-\pi / 4}^{\pi / 4} \cos t d t \geq \pi \sqrt{2} / 4\)

Step-by-Step Solution

Answer

Yes, the inequality holds: \( \int_{-\pi/4}^{\pi/4} \cos t \, dt = \sqrt{2} \geq \frac{\pi \sqrt{2}}{4} \).

Step 1: Understand the Function

We want to compare the integral of \( \cos t \) over the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \) to a constant value, \( \frac{\pi \sqrt{2}}{4} \). The function \( \cos t \) is positive and continuous on this interval.

Step 2: Evaluate Potential Comparisons

Since \( \cos t \) is a known trigonometric function, let's consider its maximum and look for a simple comparison function. Notice that \( \cos t \geq \frac{\sqrt{2}}{2} \) over the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \), because \( \cos \left( \pm \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \).

Step 3: Set Up the Inequality

According to the comparison theorem, \( \int_{a}^{b} f(t) \, dt \geq \int_{a}^{b} g(t) \, dt \) if \( f(t) \geq g(t) \) for all \( t \) in \([a, b]\). Thus, we have \( \cos t \geq \frac{\sqrt{2}}{2} \) on the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \).

Step 4: Perform the Integration

Now integrate both functions over the interval: \( \int_{-rac{\pi}{4}}^{\frac{\pi}{4}} \cos t \, dt \geq \int_{-rac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sqrt{2}}{2} \, dt \). Evaluating these, we have:- \( \int_{-rac{\pi}{4}}^{\frac{\pi}{4}} \cos t \, dt = [\sin t]_{-rac{\pi}{4}}^{\frac{\pi}{4}} = \sin(\frac{\pi}{4}) - (-\sin(\frac{\pi}{4})) = \sqrt{2}/2 + \sqrt{2}/2 = \sqrt{2} \).- \( \int_{-rac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sqrt{2}}{2} \, dt = \frac{\sqrt{2}}{2} \times \frac{\pi}{2} = \frac{\pi \sqrt{2}}{4} \).

Step 5: Conclusion

From the above calculations, \( \sqrt{2} \geq \frac{\pi \sqrt{2}}{4} \) or equivalently \( \int_{-rac{\pi}{4}}^{\frac{\pi}{4}} \cos t \, dt \geq \frac{\pi \sqrt{2}}{4} \) is true, confirming the problem statement using the comparison theorem.

Key Concepts

Trigonometric functions

Trigonometric functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. The most common trigonometric functions are sine (\( \sin \theta \)), cosine (\( \cos \theta \)), and tangent (\( \tan \theta \)). These functions are essential in understanding oscillatory and wave-like phenomena.
In the exercise, we specifically deal with the cosine function, \( \cos t \). This function is periodic with a period of \( 2\pi \), meaning it repeats every \( 2\pi \) interval. Over the interval \([-\frac{\pi}{4}, \frac{\pi}{4}]\), the cosine function starts at a value of \(\frac{\sqrt{2}}{2}\), reaches a maximum of 1 at \( t = 0 \), and returns back to \(\frac{\sqrt{2}}{2}\).

• Cosine is positive between \(-\frac{\pi}{4}\) and \(\frac{\pi}{4}\).
• The function is symmetric around the y-axis, making it an even function. This symmetry simplifies calculations because the left and right halves of the interval cancel each other in some ways when analyzing areas under the curve.

Definite integral

The definite integral is a fundamental concept in calculus. It provides a way of finding the area under a curve over a specified interval. In mathematical notation, it is written as \( \int_{a}^{b} f(t) \, dt \).
In our exercise, we evaluate the definite integral of the cosine function over the interval \([-\frac{\pi}{4}, \frac{\pi}{4}]\) to find the area beneath the graph of \(\cos t\) between these limits. This involves finding the antiderivative of \(\cos t\), which is \(\sin t\), and then applying the limits:

• The calculation of \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos t \, dt \) results in \( \sqrt{2} \).
• The calculated integral tells us the net area bounded by the x-axis and the curve from \( t = -\frac{\pi}{4} \) to \( t = \frac{\pi}{4} \).

By summarizing the integral's result, one can see how the comparison theorem plays a role in showing that this area is greater than a specific constant value.

Inequalities in calculus

Inequalities in calculus often help us establish bounds and can serve as essential tools in approximating values within definite integrals. They can determine the relationship between functions over intervals, providing insights into the behavior of these functions without precise evaluation.
The comparison theorem in calculus allows us to compare integrals when a function is known to be greater than or equal to another over a certain interval. In our problem, the comparison theorem is used to establish that the integral of \( \cos t \) is at least as large as that of the constant function \( \frac{\sqrt{2}}{2} \) over the same interval.

• The theorem stipulates that if \( f(t) \geq g(t) \) for all \( t \) in \([a, b]\), then \( \int_{a}^{b} f(t) \, dt \geq \int_{a}^{b} g(t) \, dt \).
• In this specific case, upon evaluating both expressions, we confirm: \( \sqrt{2} \geq \frac{\pi\sqrt{2}}{4} \).

This use of inequalities allows us to accurately determine relationships between complex functions using simpler ones as benchmarks.