Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

apply integration by parts twice to evaluate each integral. $$ \int r^{2} \sin r d r $$

Step-by-Step Solution

Answer

The integral is \(-r^2 \cos r + 2r \sin r + 2 \cos r + C\).

Step 1: Identify Parts for Integration by Parts

For integration by parts, use the formula \( \int u \, dv = uv - \int v \, du \). For the given integral \( \int r^2 \sin r \, dr \), choose \( u = r^2 \) and \( dv = \sin r \, dr \). This gives us \( du = 2r \, dr \) and \( v = -\cos r \).

Step 2: Apply Integration by Parts Once

Using the chosen parts, apply the integration by parts formula: \[ \int r^2 \sin r \, dr = -r^2 \cos r + \int 2r \cos r \, dr. \] Now, we need to evaluate \( \int 2r \cos r \, dr \).

Step 3: Choose Parts for the Second Integral

For \( \int 2r \cos r \, dr \), apply integration by parts again. Choose \( u = 2r \), \( dv = \cos r \, dr \). Then \( du = 2 \, dr \) and \( v = \sin r \).

Step 4: Apply Integration by Parts a Second Time

Apply the integration by parts formula to \( \int 2r \cos r \, dr \): \[ \int 2r \cos r \, dr = 2r \sin r - \int 2 \sin r \, dr. \]

Step 5: Integrate the Remaining Simple Integral

Calculate the simpler integral \( \int 2 \sin r \, dr \): \[ \int 2 \sin r \, dr = -2 \cos r + C. \]

Step 6: Substitute Back into Earlier Steps

Substitute the result from Step 5 back into Step 4: \[ \int 2r \cos r \, dr = 2r \sin r + 2 \cos r - C. \] Then substitute this into Step 2: \[ \int r^2 \sin r \, dr = -r^2 \cos r + 2r \sin r + 2 \cos r - C. \]

Step 7: Finalize the Integration Result

Combine all the terms and simplify if possible. The constant of integration is arbitrary, so you can rewrite it as simply \( + C \). The final result is: \[ \int r^2 \sin r \, dr = -r^2 \cos r + 2r \sin r + 2 \cos r + C. \]

Key Concepts

Calculus

Calculus is a fundamental branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series. It helps us understand change and motion, much like how algebra involves the study of operations and their application to solving equations. Two main branches of calculus are differential calculus and integral calculus, each addressing different types of problems.

• Differential Calculus: This branch deals with the concept of a derivative, which represents rates of change. It is widely used in fields such as physics, engineering, and economics to understand the dynamics of systems.
• Integral Calculus: This branch revolves around the concept of integration, which is about finding accumulations, such as areas under curves or total quantities from rates of change. It integrates new knowledge over time or space.

Calculus is essentially about understanding how things change and accumulate, providing a powerful toolset for modeling and solving real-world problems.

Definite and Indefinite Integrals

Integrals are a core concept in calculus, representing the area under the curve of a graph of a function. There are two types of integrals: definite and indefinite.

• Definite Integral: It has limits, which specify the start and end points of integration. The result is a number representing the accumulated quantity, such as area, between these two points. For example, the definite integral of function \( f(x) \) from \( a \) to \( b \) is written as \( \int_{a}^{b} f(x) \, dx \). It is evaluated using the Fundamental Theorem of Calculus.
• Indefinite Integral: This integral does not have specified limits, representing a family of functions. It leads to finding the antiderivative, which is the reverse process of differentiation. It is expressed as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative and \( C \) is the constant of integration.

Indefinite integrals are often used to form general solutions to differential equations, while definite integrals are applied in calculating precise values for various scientific and engineering applications.

Advanced Integration Techniques

Integration by parts is a crucial advanced technique in calculus used to solve integrals that are products of functions and not easily simplified by basic integration rules. This technique stems from the product rule of differentiation and is used when straightforward methods don't work.

• The Formula: The integration by parts formula is given by \( \int u \, dv = uv - \int v \, du \). This formula is strategic, as choosing \( u \) and \( dv \) properly can simplify the integral substantially.
• Choosing Functions: Selecting \( u \) and \( dv \) requires practice. Generally, you pick \( u \) to be a function that becomes simpler upon differentiation, while \( dv \) is a function that can be easily integrated. This thoughtful selection makes the resulting calculations straightforward.
• Multiple Applications: As seen in the given exercise, integration by parts can be applied more than once. After the first application, you might find another integral that can be simplified using the same technique. Applying it systematically can help solve complex functions step by step.

Mastering integration by parts and similar techniques allows tackling challenging integrals, often encountered in advanced calculus topics and real-world problem-solving.