Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

, use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{1} x \sin \left(\pi x^{2}\right) d x $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{1}{\pi} \).

Step 1: Identify the substitution

We start by identifying the substitution needed to simplify the integral. Notice that inside the sine function, we have \(\pi x^2\). Thus, we set \(u = \pi x^2\). Calculate the derivative of \(u\) with respect to \(x\) which gives \(du = 2\pi x \, dx\).

Step 2: Solve for dx

From \(du = 2\pi x \, dx\), we express \(dx\) in terms of \(du\): \(dx = \frac{du}{2\pi x}\).

Step 3: Change limits of integration

When \(x = 0\), \(u = \pi (0)^2 = 0\). When \(x = 1\), \(u = \pi (1)^2 = \pi\). Therefore, the new limits of integration for \(u\) are from 0 to \(\pi\).

Step 4: Substitute into the integral

Substitute \(u\) and \(dx\) into the original integral. The integral becomes: \[\int_{0}^{\pi} \frac{x}{x} \sin(u) \frac{du}{2\pi} = \int_{0}^{\pi} \frac{1}{2\pi} \sin(u) \, du\]

Step 5: Evaluate the new integral

This integral can be directly evaluated: \[\int_{0}^{\pi} \frac{1}{2\pi} \sin(u) \, du = \left[ -\frac{1}{2\pi} \cos(u) \right]_{0}^{\pi}\]For \(u = \pi\), \(\cos(\pi) = -1\). For \(u = 0\), \(\cos(0) = 1\).

Step 6: Calculate the definite integral

Substitute the upper and lower limits into the evaluated function:\[\left[ -\frac{1}{2\pi} (-1) \right] - \left[ -\frac{1}{2\pi} (1) \right]\]Simplifying this expression, we get:\[\frac{1}{2\pi} + \frac{1}{2\pi} = \frac{2}{2\pi} = \frac{1}{\pi}\]

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, often used to determine the area under a curve between two points. Unlike indefinite integrals, which represent families of functions or antiderivatives, definite integrals result in a specific numerical value. In definite integrals, limits of integration are specified, giving a boundary for evaluation. These limits are depicted as numbers written at the lower and upper bounds of the integral symbol. For instance, \[ \int_{a}^{b} f(x) \, dx \]where \(a\) and \(b\) are the limits, defines the definite integral of the function \(f(x)\) from \(x = a\) to \(x = b\).
Using definite integrals involves several steps:

• Finding the antiderivative of the function, called the indefinite integral.
• Applying the limits of integration to this antiderivative.
• Subtracting the lower limit value from the upper limit value.

Definite integrals are useful in fields like physics and engineering for calculating quantities like displacement, area, and more.

Trigonometric Substitution

Trigonometric substitution is a calculus technique used when faced with integrals involving square roots or quadratic expressions. This method simplifies the integral by substituting a trigonometric function for a variable, effectively transforming a complex expression into an easier trigonometric integral.
When using trigonometric substitution, there are typical substitutions:

• For \( \sqrt{a^2 - x^2} \), use \( x = a \sin \theta \).
• For \( \sqrt{a^2 + x^2} \), use \( x = a \tan \theta \).
• For \( \sqrt{x^2 - a^2} \), use \( x = a \sec \theta \).

After substituting, the integral becomes a function of \( \theta \), which often has a standard solution. After evaluating, we revert the variable back using the initial substitution relationships.

Calculus Techniques

In calculus, various techniques are employed to tackle integrals, each suited to different functions. Some common methods include:

• Substitution Rule: Simplifies an integral by changing variables. We introduce a new variable, \(u\), to express the integral in terms of \(u\) rather than \(x\).
• Integration by Parts: Useful when integrating the product of two functions. This technique is based on the product rule for differentiation.
• Partial Fraction Decomposition: Used for rational functions. It involves expressing the function as a sum of simpler fractions, making the integral easier to solve.

These techniques help break down complex integrals into manageable parts or simpler forms. Understanding when and how to apply each method is crucial for solving integrals efficiently.

Integral Evaluation

Evaluating integrals is the process of finding a numerical value for a definite integral. This requires understanding the function, applying appropriate techniques, and calculating the antiderivatives.
The evaluation follows these steps:

• Identify a method for integration. This could be substitution, integration by parts, etc.
• Perform the integration by calculating the antiderivative of the transformed function.
• Apply the limits of integration to the antiderivative.
• Subtract the lower limit evaluation from the upper limit evaluation to find the result.

In the exercise we reviewed, substitution was used to transform the function into a simpler form. Then, the definite integral was evaluated using the new limits and the antiderivatives, resulting in an expression that yielded a precise numerical value. This example highlights the process of transforming and evaluating complex integrals into understandable results.