Problem 79 - Calculus for Biology and Medicine

Question

Verify each inequality without evaluating the integrals. $$ \frac{\pi}{3} \leq \int_{\pi / 6}^{5 \pi / 6} \sin x d x \leq \frac{2 \pi}{3} $$

Step-by-Step Solution

Answer

The inequality is verified as true within the given bounds.

Step 1: Understand the Inequality Problem

We are asked to verify the inequality \( \frac{\pi}{3} \leq \int_{\pi / 6}^{5 \pi / 6} \sin x \, dx \leq \frac{2\pi}{3} \) without evaluating the integral directly. This involves checking if the integral, which represents the area under the curve \( y = \sin x \) from \( x = \frac{\pi}{6} \) to \( x = \frac{5\pi}{6} \), lies within the given bounds.

Step 2: Analyze the Integral Bounds

The function \( \sin x \) varies between -1 and 1, but for \( x \) in the interval \( \left[ \frac{\pi}{6}, \frac{5\pi}{6} \right] \), it is always non-negative since it is within the first and second quadrants. The maximum value \( \sin x \) can take in this interval is 1, and the minimum value is 0.5 (at the endpoints). Thus, \( \frac{1}{2} \) \( \leq \sin x \leq 1 \) in the interval.

Step 3: Set Up the Inequalities

To find the bounds of the integral, multiply each side of the inequality \( \frac{1}{2} \leq \sin x \leq 1 \) by the width of the interval \( \left( \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \right) \). This yields: \[\frac{1}{2} \times \frac{2\pi}{3} \leq \int_{\pi / 6}^{5\pi / 6} \sin x \, dx \leq 1 \times \frac{2\pi}{3}\] Simplify the left and right bounds: \[\frac{\pi}{3} \leq \int_{\pi / 6}^{5\pi / 6} \sin x \, dx \leq \frac{2\pi}{3}\]

Step 4: Conclusion

The inequalities have been verified by analyzing the bounds of the function \( \sin x \) over the given interval. Because \( \sin x \) lies between its minimum and maximum possible values over the interval, multiplying by the interval width confirms the bounds \( \frac{\pi}{3} \) and \( \frac{2\pi}{3} \) for the integral.

Key Concepts

Understanding the Sin Function

The sine function, denoted as \( \sin(x) \), is one of the fundamental trigonometric functions. It describes a smooth, oscillating wave, with periodic behavior that repeats every \(2\pi\) units.

In the context of this exercise, understanding the sine function's behavior in the interval from \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \) is crucial. During this range, the sine function is positive and exhibits values between \(0.5\) and \(1\).

Some key characteristics of the sine function include:

• Amplitude: The maximum height from its wave's center (0) to its highest point, which is 1 for the sine function.
• Periodicity: Sine repeats every \(2\pi\) radians.
• Range: For \(x\) between \(0\) and \(\pi\), \(\sin(x)\) only delivers positive values.

Knowledge of these traits helps us verify the inequality presented without needing to evaluate the integral.

Definite Integrals Explained

A definite integral is represented as \( \int_{a}^{b} f(x) \, dx \), and it calculates the area under the curve \(f(x)\) between the limits \(a\) and \(b\). This area can be thought of as the accumulation of the function's immediate values across an interval.

When handling expressions like \( \int_{\pi / 6}^{5\pi / 6} \sin x \, dx \), we are interested in the general behavior and how this area under the sine curve confines itself between two bounds.

• Boundaries: The area is finite and bounded by the function values between \(a\) and \(b\).
• Zero Transition: The area might include regions where the function passes through zero, but in our case, it remains positive.
• Graphical Interpretation: Viewing this mathematically graph helps identify why the sine function's typical oscillation easily complies with the bounds when integrated.

Understanding these ideas provides insight into verifying integral inequalities by visualizing the contained area.

Unraveling Function Bounds

Function bounds help us frame the potential outputs of a function for specific inputs. Recognizing these bounds, especially when \(f(x)\) represents a trigonometric function like sine, can streamline verifying inequalities without direct computation of integrals.

For the sine function over the interval \( \left[ \frac{\pi}{6}, \frac{5\pi}{6} \right] \):

• Minimum Value: The function never dips below \(0.5\) in our specific interval, as both endpoints mark \(\sin(x)=0.5\).
• Maximum Value: The peak is achieved at \(\sin(x)=1\) within the studied portion.
• Bounded Integral: Multiplying the bounds \(0.5 \leq \sin x \leq 1\) by the width \(\left( \frac{4\pi}{6} \right)\) gives overall confines for the integral's results.

Accurately identifying and using these bounds aids in calculating the integral's expected range, simplifying resolution of inequalities.