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Solve each integral. Each can be found using rules developed in this section, but some algebra may be required. $$ \int(x-1)^{2} x^{3} d x $$

Short Answer

Expert verified
The integral is \( \frac{1}{6}x^6 - \frac{2}{5}x^5 + \frac{1}{4}x^4 + C \).

Step by step solution

01

Expand the Binomial

First, expand the expression \( (x-1)^2 \) using the binomial expansion formula. \[ (x-1)^2 = x^2 - 2x + 1. \] This allows us to rewrite the original integral as \[ \int (x^2 - 2x + 1) x^3 \, dx. \]
02

Simplify the Expression

Distribute \( x^3 \) through each term in the parenthesis. Thus, \[(x^2 - 2x + 1) x^3 = x^5 - 2x^4 + x^3. \] The integral now becomes \[ \int (x^5 - 2x^4 + x^3) \, dx. \]
03

Integrate Term by Term

Use the power rule for integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) where \( n \) is not equal to -1. Integrate each term separately:- \( \int x^5 \, dx = \frac{x^6}{6} \)- \( \int 2x^4 \, dx = \frac{2x^5}{5} \)- \( \int x^3 \, dx = \frac{x^4}{4} \)Combine the integrated terms: \[ \frac{x^6}{6} - \frac{2x^5}{5} + \frac{x^4}{4} + C. \]
04

Simplify the Result

Simplify the expression by writing it in a cleaner form: \[ \frac{1}{6}x^6 - \frac{2}{5}x^5 + \frac{1}{4}x^4 + C. \] This is the final result of the integration and is in the simplest form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Expansion
Binomial expansion is a crucial mathematical tool, especially in calculus. It allows us to expand expressions that are raised to a power, such as \((x-1)^2\). Imagine it as a method of breaking down a complicated power expression into a simpler sum of terms.

For instance, in the expression \((x-1)^2\), binomial expansion helps rewrite it as \(x^2 - 2x + 1\). This transformation is indispensable when dealing with integrals, as it simplifies the integrand, making it more manageable.
  • First, identify the expression that needs expansion, like \((x-1)^2\).
  • Apply the formula \((a-b)^2 = a^2 - 2ab + b^2\).
  • Rewrite the expression in an expanded form: \(x^2 - 2x + 1\).
Thus, by expanding binomials, you transform integrals into a simpler form, setting a solid foundation for easier integration.
Power Rule for Integration
Integration is a fundamental operation in calculus, and one of the easiest rules to apply is the Power Rule. This rule helps you integrate functions of the form \(x^n\). Let's break it down for clarity:
  • The power rule states: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), assuming \(n eq -1\).
  • Here's how to use it: take the exponent \(n\), add 1 to it, then divide the term by this new exponent. Don't forget the constant \(C\), which represents the indefinite integral's constant of integration.
Let's apply it practically. For instance, with \( \int x^5 \, dx\), you'll increase the exponent to 6 and divide by 6, resulting in \(\frac{x^6}{6}\).

Using the power rule makes integration straightforward and is especially useful for polynomial functions. Each term can be handled individually, simplifying complex integrals one step at a time.
Definite and Indefinite Integrals
Understanding the difference between definite and indefinite integrals is essential in calculus.

An indefinite integral, like the one we are dealing with, \(\int (x^5 - 2x^4 + x^3) \, dx\), doesn't have limits of integration. Its result is a family of functions, including a constant \(C\). You can see it as the opposite operation of differentiation, finding the original function from its derivative.

On the other hand, definite integrals are evaluated over a specific interval \([a,b]\). They calculate the net area under the curve \(y = f(x)\) from \(x = a\) to \(x = b\). The result is a number, giving the accumulation of quantities, like total distance or area.
  • Indefinite Integrals: No limits. Result includes \(C\).
  • Definite Integrals: Calculated over an interval \([a,b]\). Result is a specific value.
Understanding these fundamentals of integration will enable you to solve a wide range of real-world problems in mathematics and science.

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