Problem 92 - Calculus: Early Transcendentals

Question

If \( f \) is continuous on \( [0, \pi] \), use the substitution \( u = \pi - x \) to show that $$ \int^{\pi}_0 x f(\sin x) \,dx = \frac{\pi}{2} \int^{\pi}_0 f(\sin x) \,dx $$

Step-by-Step Solution

Answer

Substitute \( u = \pi - x \) to balance the integral and exploit symmetry to show equality.

Step 1: Identify the Substitution

We start by substituting using the given transformation. Let \( u = \pi - x \). This implies that the derivative \( du = -dx \). Also, the limits of integration change: when \( x = 0 \), \( u = \pi \) and when \( x = \pi \), \( u = 0 \).

Step 2: Rewrite the Integral with New Variable

We rewrite the original integral in terms of \( u \). Using the substitution \( u = \pi - x \), the original integral becomes \[ \int_{\pi}^{0} (\pi - u) f(\sin(\pi - u)) (-du). \] Changing bounds switches the sign, so we write: \[ \int_{0}^{\pi} (\pi - u) f(\sin u) \, du. \]

Step 3: Simplify the Substituted Integral

The expression \( \sin(\pi - u) \) simplifies to \( \sin u \) due to the sine addition formula. So, the transformed integral becomes \[ \int_{0}^{\pi} (\pi - u) f(\sin u) \, du. \]

Step 4: Break Down the Simplified Integral

Separate the integral into parts: \[ \int_{0}^{\pi} \pi f(\sin u) \, du - \int_{0}^{\pi} u f(\sin u) \, du. \] The first integral is \[ \pi \int_{0}^{\pi} f(\sin u) \, du, \] and the second is \[ \int_{0}^{\pi} u f(\sin u) \, du. \]

Step 5: Internal Symmetry Result

Notice that \( \int_{0}^{\pi} x f(\sin x) \, dx \) is equal to \( \int_{0}^{\pi} u f(\sin u) \, du \) by symmetry. Thus the original problem becomes a balance of the two expressions.

Step 6: Equate and Simplify

Realize that \[ 2 \int_{0}^{\pi} x f(\sin x) \, dx = \pi \int_{0}^{\pi} f(\sin x) \, dx, \] which simplifies to \[ \int_{0}^{\pi} x f(\sin x) \, dx = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx. \] This satisfies the condition we are proving.

Key Concepts

Substitution Method

The Substitution Method is a powerful tool in integral calculus. It helps simplify integrals by changing variables. This technique often makes complex integrals more manageable. Here, we transform an inconvenient function into a simpler form by substituting variables. The key steps involve:

• Selecting an appropriate substitution, such as the provided one: \( u = \pi - x \).
• Finding the differential, i.e., \( du = -dx \). This tells us how the small changes in \( u \) relate to those in \( x \).
• Changing the limits of integration according to the new variable. For instance, when \( x = 0 \), \( u = \pi \) and when \( x = \pi \), \( u = 0 \).
• Rewriting the integral in terms of the new variable, which often results in a more straightforward integral to evaluate.

Utilizing these systematic steps, we can tackle otherwise tough integral problems efficiently.

Definite Integrals

Definite Integrals compute the signed area under a curve over a specified interval. They provide valuable information about a function between two bounds. In this exercise, we deal with the integral from \( 0 \) to \( \pi \).For definite integrals, remember:

• They have limits of integration, such as \([0, \pi]\) in this case.
• The result depends heavily on the behavior of the function within these limits.
• The fundamental theorem of calculus often comes into play, linking derivatives and integrals.
• Switching the limits of the integral reverses the sign.
• They help in finding values like total distance, area, or accumulated quantity over a range.

Through definite integrals, this problem helps demonstrate the symmetry properties of the given function across the specified interval.

Continuous Functions

Continuous Functions are integral-friendly because they have no breaks or holes over their domain. Their smooth nature ensures the existence of definite integrals. In our exercise, the function \( f \) is continuous on \([0, \pi]\).Characteristics of continuous functions include:

• They are predictable and do not jump from one value to another abruptly.
• Small changes in \( x \) lead to small changes in \( f(x) \).
• Continuous functions guarantee that the limit of the function as \( x \) approaches any point equals the value of the function at that point.
• This makes evaluating integrals straightforward because there are no sudden breaks or infinite values.

In calculus, dealing with continuous functions ensures that several mathematical properties, such as secant line approximation and definite integral computation, hold true seamlessly.

Symmetry in Integrals

Symmetry in Integrals often simplifies calculations by recognizing patterns within the function over a given interval. In the original problem, symmetry plays a crucial role.Understanding symmetry can involve:

• Identifying if a function is even or odd. This helps simplify integrals over symmetric intervals like \([-a, a]\).
• For functions symmetric around a point, integrals can often be rearranged or simplified.
• In this problem, the symmetry of \( f(\sin x) \) across the interval \([0, \pi]\) lets us equate different parts of the integral to simplify it significantly.
• Symmetric properties can lead to results where calculations are reduced, such as showing that \( \int_0^\pi x f(\sin x) \, dx = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx \).

Recognizing and using symmetry can save a lot of time and effort in integral calculations, turning potentially complex problems into elegant solutions.