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Magazine Chapter 5: Problem 45
Evaluate the integrals. $$\int(x+1)^{2}(1-x)^{5} d x$$
Short Answer
Expert verified
The integral is \( \frac{32}{3}(x+1)^3 - 20(x+1)^4 + 16(x+1)^5 - \frac{20}{3}(x+1)^6 + \frac{10}{7}(x+1)^7 - \frac{1}{8}(x+1)^8 + C \).
Step by step solution
01
Use Substitution Method
We begin by using the substitution method to simplify the integral. Let \( u = x + 1 \). Then, \( du = dx \) because the derivative of \( x+1 \) with respect to \( x \) is 1. We also need to express \( 1-x \) in terms of \( u \). We have \( x = u - 1 \), so \( 1-x = 1-(u-1) = 2-u \). This changes the integral to: \[ \int u^2 (2-u)^5 \, du. \]
02
Expand the Expression
Expand \((2-u)^5\) using the binomial theorem or by repeated application of the distributive property. When you expand, you get the polynomial:\[ (2-u)^5 = \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (-u)^k = 32 - 80u + 80u^2 - 40u^3 + 10u^4 - u^5. \]
03
Multiply Polynomials
Multiply \( u^2 \) by each term in the expanded form of \( (2-u)^5 \). This results in:\[ \int (32u^2 - 80u^3 + 80u^4 - 40u^5 + 10u^6 - u^7) \, du. \]
04
Integrate Term by Term
Now integrate each term separately:- \( \int 32u^2 \, du = \frac{32}{3}u^3 \)- \( \int -80u^3 \, du = -20u^4 \)- \( \int 80u^4 \, du = 16u^5 \)- \( \int -40u^5 \, du = -\frac{40}{6}u^6 = -\frac{20}{3}u^6 \)- \( \int 10u^6 \, du = \frac{10}{7}u^7 \)- \( \int -u^7 \, du = -\frac{1}{8}u^8 \)Combine these to get the antiderivative:\[ \frac{32}{3}u^3 - 20u^4 + 16u^5 - \frac{20}{3}u^6 + \frac{10}{7}u^7 - \frac{1}{8}u^8 + C, \] where \( C \) is the constant of integration.
05
Substitute Back in Terms of x
Replace \( u \) with \( x+1 \) to express the solution in terms of \( x \):\[ \frac{32}{3}(x+1)^3 - 20(x+1)^4 + 16(x+1)^5 - \frac{20}{3}(x+1)^6 + \frac{10}{7}(x+1)^7 - \frac{1}{8}(x+1)^8 + C. \]This is the integrated expression in terms of \( x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a fundamental tool for solving integrals. It simplifies complex integrals by changing variables, which can make the integral easier to evaluate. In this method:
- Identify a part of the integral you can substitute with a variable, usually one that simplifies the expression.
- Let this part be equal to a new variable, say, \( u \). Write down the corresponding \( du \) by differentiating the substituted variable with respect to the original variable.
- Express the entire integral in terms of \( u \) and \( du \).
Polynomial Expansion
Polynomial expansion involves expressing expressions raised to a power as a sum of simpler terms. This step is often necessary in integration problems like the exercise at hand, where dealing with a reduced form can aid integration. Here's how this expansion works:
- Take a polynomial, like \((2-u)^5\), and apply methods such as the distributive property or known identity expansions like the binomial theorem.
- Expand it into a sum of terms, such as \(32 - 80u + 80u^2 - 40u^3 + 10u^4 - u^5\) in our example.
Binomial Theorem
The binomial theorem is a powerful algebraic tool that provides a formula for expanding expressions of the form \((a + b)^n\). This theorem is especially useful when expanding expressions within integrals to simplify calculations.
- The binomial theorem states: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \). Here, \( \binom{n}{k} \) is the binomial coefficient, representing combinations of \( n \) items taken \( k \) at a time.
- Using it, \((2-u)^5\) is expanded as \( \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} (-u)^k \), resulting in the series of terms seen in our exercise.
