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Chapter 5: Problem 33

Evaluate the integral. $$ \int_{0}^{1}(1+r)^{3} d r $$

Short Answer

Expert verified
The value of the integral is \( \frac{15}{4} \).

Step by step solution

01

Expand the Integrand

First, expand the function \((1 + r)^3\) using the Binomial Theorem. The expansion is: \[(1 + r)^3 = 1 + 3r + 3r^2 + r^3.\]
02

Write the Expanded Integral

After expanding, the integral becomes \[\int_{0}^{1} (1 + 3r + 3r^2 + r^3) \, dr.\] This separates into four simpler integrals:\[\int_{0}^{1} 1 \, dr + \int_{0}^{1} 3r \, dr + \int_{0}^{1} 3r^2 \, dr + \int_{0}^{1} r^3 \, dr.\]
03

Integrate Each Term

Integrate each term separately:- The integral of \(1\) is \(r\): \[\int_{0}^{1} 1 \, dr = [r]_0^1 = 1.\]- The integral of \(3r\) is \(\frac{3r^2}{2}\): \[\int_{0}^{1} 3r \, dr = \left[\frac{3r^2}{2}\right]_0^1 = \frac{3}{2}.\]- The integral of \(3r^2\) is \(r^3\): \[\int_{0}^{1} 3r^2 \, dr = [r^3]_0^1 = 1.\]- The integral of \(r^3\) is \(\frac{r^4}{4}\): \[\int_{0}^{1} r^3 \, dr = \left[\frac{r^4}{4}\right]_0^1 = \frac{1}{4}.\]
04

Combine the Results

Add the results of the four integrals:\[1 + \frac{3}{2} + 1 + \frac{1}{4} = \frac{4}{4} + \frac{6}{4} + \frac{4}{4} + \frac{1}{4} = \frac{15}{4}.\]

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Key Concepts

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

Binomial Theorem
The Binomial Theorem provides a powerful way to expand powers of binomials. A binomial is simply a two-term expression, like (1 + r). The theorem describes how to expand it when it is raised to a power, such as (1 + r)^3, by using a formula that involves coefficients from Pascal's triangle.

For example, according to the Binomial Theorem:
  • (1 + r)^3 = 1 + 3r + 3r^2 + r^3
These coefficients (1, 3, 3, 1) come from the fourth row of Pascal's triangle. It's useful for breaking down expressions into simpler parts, which can then be integrated individually as they appear in definite integrals.

The expansion simplifies complex polynomial integration tasks and is a crucial step before applying integration techniques.
Integration Techniques
Integration techniques help us solve integrals, which are the reverse operation of derivatives. In particular, definite integrals like \(\int_{a}^{b} f(r) \, dr\) give us the signed area under the curve from r = a to r = b.

After using the Binomial Theorem, the integral
\(\int_{0}^{1}(1 + 3r + 3r^2 + r^3) \, dr\)can be broken down into simpler integrals.
  • \(\int_{0}^{1} 1 \, dr\)
  • \(\int_{0}^{1} 3r \, dr\)
  • \(\int_{0}^{1} 3r^2 \, dr\)
  • \(\int_{0}^{1} r^3 \, dr\)
This step not only simplifies the process but also allows each term to be easily integrated, using basic rules for polynomial integration such as the power rule.
Polynomial Integration
Polynomial integration involves integrating each term of a polynomial separately. The polynomial is formed by adding up terms which have variables raised to powers, like 3r^2. Luckily, integrating a polynomial term is straightforward with the power rule.

The power rule states that:\(\int r^n \, dr = \frac{r^{n+1}}{n+1} + C\)where C is the constant of integration, which is not needed when working with definite integrals.
  • The integral of 3r becomes \(\frac{3r^2}{2}\) after evaluating.
  • The integral of 3r^2 simplifies to r^3.
  • The term r^3 integrates to \(\frac{r^4}{4}\).
By applying the power rule to each term and evaluating from the specified limits, you find the definite integral's value. For this problem, after integration, summing these results gives \(\frac{15}{4}\).

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