Problem 45 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int(x+1)^{2}(1-x)^{5} d x$$

Step-by-Step Solution

Answer

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 1: 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: \[ \int u^2 (2-u)^5 \, du. \]

Step 2: 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:\[ (2-u)^5 = \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (-u)^k = 32 - 80u + 80u^2 - 40u^3 + 10u^4 - u^5. \]

Step 3: Multiply Polynomials

Multiply \( u^2 \) by each term in the expanded form of \( (2-u)^5 \). This results in:\[ \int (32u^2 - 80u^3 + 80u^4 - 40u^5 + 10u^6 - u^7) \, du. \]

Step 4: 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.

Step 5: 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 \( x \).

Key Concepts

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 \( du \).

For the given exercise, we set \( u = x + 1 \), making \( du = dx \). This changes our original integral to a simpler form. The goal is to evaluate the integral in terms of \( u \), and finally substitute back to return to the original variable, \( x \). This method helps in breaking down cumbersome integration processes and is particularly useful when handling polynomials and chain rule derivatives.

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.

Breaking down powers of binomials into individual terms allows for straightforward integration of each term individually. This step prepares the groundwork for the next steps where each term from the expansion is integrated separately. Polynomial expansion transforms challenging expressions into a series of simpler, manageable terms.

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.

This expansion uses both algebraic manipulation and an understanding of combinations. It simplifies integration by reducing a complex expression into a series of terms, each of which can be integrated separately. The binomial theorem is crucial for handling expressions in calculus efficiently.