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Magazine Chapter 6: Problem 88
After reading the preceding example, find each integral by repeated integration by parts using a table. $$ \int(x+1)^{2}(x+2)^{5} d x $$
Short Answer
Expert verified
Integral: \( \frac{(x+1)^2 (x+2)^6}{6} - \frac{2}{6} \left( \frac{(x+1)(x+2)^7}{7} - \frac{(x+2)^8}{56} \right) + C \).
Step by step solution
01
Recognize the Integration by Parts Formula
The integration by parts formula is given by \( \int u \, dv = uv - \int v \, du \). We will use this formula repeatedly, which means we'll continue to apply integration by parts to each subsequent integral until we reach an elementary result.
02
Choose \( u \) and \( dv \) for Integration by Parts
For repeated integration by parts, choose \( u = (x+1)^2 \) and \( dv = (x+2)^5 \, dx \). The derivative \( du = 2(x+1) \, dx \), and we need to integrate \( dv \) to find \( v \).
03
Integrate \( dv \) to Find \( v \)
To find \( v \), integrate \( dv: \int (x+2)^5 \, dx \). Using the power rule for integration, \[ v = \frac{(x+2)^6}{6} + C \], where \( C \) is the constant of integration. For integration by parts, we'll ignore the constant momentarily.
04
Apply Integration by Parts Formula
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula, \( \int u \, dv = uv - \int v \, du \). Thus, \[ \int (x+1)^2 (x+2)^5 \, dx = \frac{(x+1)^2 (x+2)^6}{6} - \int \frac{(x+2)^6}{6} \, 2(x+1) \, dx. \] Simplify it to \[ \frac{(x+1)^2 (x+2)^6}{6} - \frac{2}{6} \int (x+2)^6 (x+1) \, dx. \]
05
Set Up a Table for Repeated Integration by Parts
Create a table to find the next \( u \) and \( dv \) for the new integral \( \int (x+2)^6 (x+1) \, dx \). Choose \( u = (x+1) \) and \( dv = (x+2)^6 \, dx \). Find their derivatives and integrals respectively.
06
Fill in the Table and Apply Formula Again
The table will help you keep track of each component:- \( u = (x+1) \)- \( dv = (x+2)^6 \, dx \) - \( du = 1 \, dx \)- \( v = \frac{(x+2)^7}{7} \)Then apply the formula:\[ \int u \, dv = uv - \int v \, du = \frac{(x+1)(x+2)^7}{7} - \int \frac{(x+2)^7}{7} \, dx. \]
07
Integrate the New Integral
Finally, integrate the leftover integral \( \int \frac{(x+2)^7}{7} \, dx \). It simplifies to \( \frac{(x+2)^8}{56} \).
08
Assemble the Final Answer
Substituting all components back gives our integral solution:\[ \int (x+1)^{2}(x+2)^{5} \, dx = \frac{(x+1)^2 (x+2)^6}{6} - \frac{2}{6} \left( \frac{(x+1)(x+2)^7}{7} - \frac{(x+2)^8}{56} \right) + C \] where \( C \) is the constant of integration. Simplify this expression to obtain the final answer.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Calculus
Calculus is a branch of mathematics that focuses on understanding changes. It primarily deals with two concepts: derivatives and integrals. Derivatives measure how a quantity changes, while integrals accumulate quantities. In this context, we are concerned with integrals, more specifically, a type of integral called a definite integral.
Calculus plays a vital role in many scientific disciplines, from physics to engineering. It enables us to solve complex problems involving rates of change and accumulation. For example:
Calculus plays a vital role in many scientific disciplines, from physics to engineering. It enables us to solve complex problems involving rates of change and accumulation. For example:
- In physics, it helps calculate object trajectories.
- In chemistry, calculus measures reaction rates.
- In economics, it models changes in market trends.
Definite Integrals
Definite integrals help us find the total accumulation of a quantity over a specific interval. They provide a numerical result, unlike indefinite integrals, which yield a formula. To compute a definite integral, you need an integrand function proposed on a specific interval
A standard integral provides accumulated values, but a definite integral specifies an upper and lower limit:
- The upper and lower limits define the interval for integration.
- It evaluates to a numerical value, representing accumulated quantity.
- Real-world applications include computing area under curves, volumes, and total accumulated quantities.
- They're fundamental for real-life problems requiring exact total measurements.
- They link calculus to practical applications such as engineering tasks.
Mathematical Formulas
Mathematical formulas play a critical role in calculus, helping solve and simplify complex problems. They provide systematic steps to follow. In integration by parts, we use a well-defined formula:\[\int u \, dv = uv - \int v \, du\]This formula comes from the derivative product rule. It allows us to integrate products of functions by differentiating and integrating different parts. The process of integration by parts often involves:
- Identifying parts \( u \) and \( dv \). The choice is crucial for simplification.
- Computing \( du \) and \( v \) from \( u \) and \( dv \).
- Substituting into the integration by parts formula.
- Simplifying and repeating the steps if necessary until an elementary result is achieved.
