Problem 70 - Calculus: Early Transcendentals

Question

(a) If \( f \) is continuous on \( [a, b] \), show that $$ \biggl| \int^b_a f(x) \,dx \biggr| \le \int^b_a \bigl| f(x) \bigr| \,dx $$ [\( Hint: \) \( -\bigl| f(x) \bigr| \le f(x) \le \bigl| f(x) \bigr|. \)] (b) Use the result of part (a) to show that $$ \biggl| \int^{2\pi}_0 f(x) \sin 2x \,dx \biggr| \le \int^{2\pi}_0 \bigl| f(x) \bigr| \,dx $$

Step-by-Step Solution

Answer

The integral of a function is bounded by the integral of its absolute value.

Step 1: Basic Understanding of Inequality

To show \( \left| \int_a^b f(x) \,dx \right| \le \int_a^b \left| f(x) \right| \,dx \), we must understand that the absolute value of an integral can be bounded by the integral of the absolute value of the function. The hint given is \( -\left| f(x) \right| \le f(x) \le \left| f(x) \right| \), which helps establish the inequality.

Step 2: Applying the Hint

Since \( f(x) \) is bounded by its absolute value as \( -\left| f(x) \right| \le f(x) \le \left| f(x) \right| \), we integrate across \([a, b]\) to get: \[ \int_a^b -\left| f(x) \right| \le \int_a^b f(x) \le \int_a^b \left| f(x) \right|. \] Thus, the range of \( \int_a^b f(x) \) is limited to the bounds set by the integral of its absolute value.

Step 3: Understanding Absolute Value of Integral

The inequality from Step 2 shows that the integral of \( f(x) \) is between \(-\int_a^b \left| f(x) \right| \) and \( \int_a^b \left| f(x) \right| \), meaning \( \left| \int_a^b f(x) \right| \le \int_a^b \left| f(x) \right| \) as it cannot surpass the bounds given by its absolute integral.

Step 4: Application of (a) to Part (b)

Using the established result from part (a), we can now address part (b). For the function \( f(x) \sin 2x \), apply the inequality: \[ \left| \int_0^{2\pi} f(x) \sin 2x \,dx \right| \le \int_0^{2\pi} \left| f(x) \sin 2x \right| \,dx. \]

Step 5: Simplification Using \( \left| \sin 2x \right| \le 1 \)

Since \( \left| \sin 2x \right| \le 1 \), we can simplify: \[ \int_0^{2\pi} \left| f(x) \sin 2x \right| \,dx \le \int_0^{2\pi} \left| f(x) \right| \,dx. \] Thus, proving \( \left| \int_0^{2\pi} f(x) \sin 2x \,dx \right| \le \int_0^{2\pi} \left| f(x) \right| \,dx. \)

Key Concepts

Continuity of Functions

In mathematical analysis, a function is continuous on an interval if it is smooth and unbroken. This means that you can draw the graph of the function on that interval without lifting your pen. Continuity is an important property when dealing with integrals because it ensures that functions behave nicely across intervals.
For the problem at hand, the continuity of the function \( f \) on \([a, b]\) guarantees that the integral \( \int_a^b f(x) \,dx \) exists, and allows us to make comparisons with the integral of its absolute value.
When a function is continuous, its interruptions or jumps are nonexistent. This is why continuous functions are particularly useful in calculus—they allow for neat integration processes. Without continuity, certain properties of these functions, such as the integral inequality being explored in our exercise, might not hold true.

Absolute Value

Absolute value is a concept used to express the magnitude of a real number, disregarding its sign. For instance, the absolute value of both \(3\) and \(-3\) is \(3\). This idea is extended to functions as well.
The absolute value function, denoted as \( |f(x)| \), measures how far \( f(x) \) is from zero, without considering the direction on the number line. This is especially crucial in integral inequalities, as seen in our exercise.
In such problems, the integral of the absolute value \( \int_a^b |f(x)| \,dx \) serves as an upper bound for the absolute value of the integral \( \left| \int_a^b f(x) \,dx \right| \). This relationship is key because it simplifies the task of comparing integrals, allowing us to establish inequalities based on the properties of absolute value.

Bounded Functions

A function is considered bounded on an interval if there exists a real number that serves as a ceiling, ensuring the function does not surpass a certain finite value across that interval. To maintain boundaries for our analysis, we require functions to remain within these upper and lower limits.
In the context of our exercise, we use the fact that \( f(x) \) is bounded by its absolute value. More specifically, we have the inequality \( -|f(x)| \leq f(x) \leq |f(x)| \).
This bound simplifies verifying integral inequalities because \( \, |f(x)| \, \) encompasses all possible values of \( f(x) \, \) over the evaluated range. This means we can manage the integral's range behavior using these bounded limits for easier analysis and outcomes.

Trigonometric Integrals

Trigonometric integrals are integrals involving trigonometric functions like sine and cosine. These integrals often arise in calculus problems, especially where oscillating functions need to be incorporated.
In our particular exercise, we dealt with an integral containing the sine function, \( \sin 2x \). Its absolute value satisfies the simple inequality \( |\sin 2x| \leq 1 \). This understanding allows us to "control" the size of expressions involving sine during integration.
Since trigonometric functions have fixed ranges, such as the range of the sine and cosine functions being \([-1, 1]\), they help us determine potential upper bounds during calculations. In applying these bounds, we can simplify integrals like \( \int_0^{2\pi} |f(x)\sin 2x| \,dx \) by recognizing that the effect of the trigonometric function will never exceed these known limits.