Problem 27 - Applied Calculus

Question

Find the integrals in problems. Check your answers by differentiation. $$ \int x \sin \left(4 x^{2}\right) d x $$

Step-by-Step Solution

Answer

The integral is \(-\frac{1}{8} \cos(4x^2) + C\).

Step 1: Identify the Integration Technique

We need to integrate the function \(x \sin(4x^2)\). This requires using integration by substitution as the integral contains a composite function. The goal is to simplify the integral by changing variables.

Step 2: Choose a Substitution

For integration by substitution, choose \(u = 4x^2\). Consequently, \( \frac{du}{dx} = 8x \), or \( du = 8x \, dx \). Therefore, \( x \, dx = \frac{1}{8} du \).

Step 3: Perform the Substitution

Substitute \(u = 4x^2\) and \(x \, dx = \frac{1}{8} du\) into the integral. It becomes: \[\int x \sin(4x^2) \, dx = \int \sin(u) \cdot \frac{1}{8} \, du = \frac{1}{8} \int \sin(u) \, du\]

Step 4: Integrate with Respect to \(u\)

The integral \(\frac{1}{8} \int \sin(u) \, du\) can be solved as follows: \[\frac{1}{8} \left(-\cos(u)\right) + C = -\frac{1}{8} \cos(u) + C\]where \(C\) is the integration constant.

Step 5: Substitute Back in Terms of \(x\)

Replace \(u\) with \(4x^2\) to express the antiderivative in terms of \(x\):\[-\frac{1}{8} \cos(4x^2) + C\] is the integral of \(x \sin(4x^2)\).

Step 6: Differentiate to Check Answer

Differentiate \(-\frac{1}{8} \cos(4x^2) + C\) with respect to \(x\) to verify:\[\frac{d}{dx} \left(-\frac{1}{8} \cos(4x^2)\right) = \frac{1}{8} \sin(4x^2) \cdot 8x = x \sin(4x^2)\]The differentiation matches the original integrand, confirming the solution.

Key Concepts

Calculus

Calculus is the branch of mathematics that allows us to study rates of change and areas under curves. It is critical for solving a range of problems in science and engineering. The main focuses of calculus are differentiation, which studies rates of change, and integration, which finds total amounts given rates of change.

When dealing with calculus, it's important to understand that:

• Integration is essentially the reverse operation of differentiation.
• The two fundamental tools of calculus are derivatives and integrals.
• Applications of calculus include finding the area under curves, optimizing solutions, and solving differential equations.

In our particular problem, we are applying integral calculus, focusing on definite integrals and how substitution simplifies complex integrals.

Definite Integrals

Definite integrals help us find the exact value of an area under a curve within a certain interval. They have limits at the top and bottom of the integral symbol, which denote the interval over which the function is being integrated.

Key aspects of definite integrals include:

• Providing the total area between the curve and the x-axis over a specific range.
• Being represented as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the points of evaluation.
• Being evaluated using the Fundamental Theorem of Calculus, which connects differentiation and integration.

In the problem we reviewed, while it's an indefinite integral example, the understanding of definite integrals lays the groundwork for more complex applications involving specific limits.

Trigonometric Functions

Trigonometric functions are crucial in calculus as they often appear in diverse mathematical problems involving periodic phenomena. The primary trigonometric functions are sine, cosine, and tangent, which relate angles to ratios of sides in a right triangle.

Applying trigonometric functions in calculus involves:

• Recognizing common trigonometric identities that simplify expressions.
• Integrating and differentiating trigonometric functions using standard formulas.
• Understanding their periodic nature and how it affects the integration and differentiation.

In the exercise given, the function involved is sine, thus leveraging the integral and derivative of sine functions was key in solving the integral once substitution was done.

Differentiation

Differentiation is finding the derivative of a function, which represents the rate of change. It is a fundamental tool in calculus, allowing us to analyze how functions change.

Important points about differentiation include:

• It calculates the slope of a function at any point.
• Rules such as the product rule, quotient rule, and chain rule facilitate finding derivatives of complex functions.
• In calculus, differentiation often verifies the results of integration.

In the example problem, differentiation was used to verify the integral solution derived by substitution. Taking the derivative of \(-\frac{1}{8} \cos(4x^2) + C\) and finding it matched the integrand, confirmed our solution was correct.