Problem 71 - Thomas Calculus

Question

Use the inequality \(\sin x \leq x,\) which holds for \(x \geq 0,\) to find an upper bound for the value of \(\int_{0}^{1} \sin x d x\)

Step-by-Step Solution

Answer

The upper bound for \(\int_{0}^{1} \sin x \, dx\) is \(\frac{1}{2}\).

Step 1: Understanding the Integral and Inequality

We are given the task of finding an upper bound for the integral \(\int_{0}^{1} \sin x \, dx\). The inequality \(\sin x \leq x\) for \(x \geq 0\) suggests we can use it to replace \(\sin x\) with its upper bound \(x\), simplifying the integral.

Step 2: Set Up the Bounding Integral

Since \(\sin x \leq x\) for \(x \geq 0\), we can state that \(\int_{0}^{1} \sin x \, dx \leq \int_{0}^{1} x \, dx\). This sets up a comparison between the original integral and a simpler integral where \(\sin x\) is replaced with \(x\).

Step 3: Evaluate the Bounding Integral

Next, we evaluate \(\int_{0}^{1} x \, dx\). This is a basic definite integral. The antiderivative of \(x\) is \(\frac{x^2}{2}\).

Step 4: Calculate the Definite Integral

Substitute the limits of integration into the antiderivative: \[\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.\]

Step 5: Conclusion of Upper Bound

Since we've calculated that \(\int_{0}^{1} x \, dx = \frac{1}{2}\), it follows that \(\int_{0}^{1} \sin x \, dx \leq \frac{1}{2}\). Hence, \(\frac{1}{2}\) is an upper bound for \(\int_{0}^{1} \sin x \, dx\).

Key Concepts

Inequality in Calculus

In calculus, inequalities are essential tools for comparing and estimating the values of functions. An inequality provides a way to understand how one function behaves in relation to another. In the exercise, we use the inequality \( \sin x \leq x \) for \( x \geq 0 \). This tells us that the sine function is always less than or equal to the linear function \( x \) when \( x \) is non-negative.
This type of inequality is useful because it allows us to simplify complex problems by replacing a complicated function with an easier, related function that serves as an upper or lower bound. Here, the inequality helps us find an upper bound for the integral of \( \sin x \). By integrating \( x \) instead, we gain important insight into the behavior of \( \sin x \) on the interval from 0 to 1.

Definite Integrals

A definite integral is a fundamental concept in calculus that represents the accumulation of quantities, which can be seen graphically as the area under a curve. In this exercise, we need to evaluate the definite integral \( \int_{0}^{1} \sin x \, dx \). However, direct computation can be challenging. Using \( \sin x \leq x \), we instead analyze \( \int_{0}^{1} x \, dx \).
To compute this integral, we find the antiderivative of the function \( x \), which is \( \frac{x^2}{2} \). We then evaluate it at the limits of integration, 0 and 1. This yields \( \frac{1}{2} \), providing the area under the line \( y = x \) from 0 to 1.

• Definite integrals have upper and lower limits, defining the range of integration.
• They calculate the net area between the function and the x-axis over the specified interval.
• These integrals offer precise information about the sum of infinite infinitesimally small quantities.

Comparison of Functions

Comparing functions is a critical aspect of analysis in calculus. By comparing the function \( \sin x \) against \( x \), as in this exercise, we find relationships that aid in solving problems more efficiently.
Here, the comparison helps by showing that the challenging integral \( \int_{0}^{1} \sin x \, dx \) is less than or equal to a simpler one, \( \int_{0}^{1} x \, dx \). This allows the problem of evaluating an unknown integral to be linked to the one with a known, simple result \( \frac{1}{2} \).
Such comparisons can also reveal other properties, such as:

• Monotonicity: Determining if a function consistently increases or decreases.
• Boundaries: Estimating limits on a function's behavior over specific intervals.
• Simplifications: Changing complex functions into more manageable forms with established results.

This method is a powerful way to understand and evaluate complex integrals by relying on simpler, known integrals.