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Chapter 5: Problem 24

Evaluate the integrals in Exercises \(1-28\). $$\int_{1}^{8} \frac{\left(x^{1 / 3}+1\right)\left(2-x^{2 / 3}\right)}{x^{1 / 3}} d x$$

Short Answer

Expert verified
The integral evaluates to 2.4.

Step by step solution

01

Expand the integrand

We begin by expanding the integrand. Multiply the expressions in the numerator:\[ (x^{1/3} + 1)(2 - x^{2/3}) = 2x^{1/3} + 2 - x^{2/3}x^{1/3} - x^{2/3}. \]This expands to:\[ 2x^{1/3} + 2 - x - x^{2/3}. \]
02

Simplify the fraction

The integrand is:\[ \frac{2x^{1/3} + 2 - x - x^{2/3}}{x^{1/3}}. \]Simplify by dividing each term in the numerator by \(x^{1/3}\):\[ \frac{2x^{1/3}}{x^{1/3}} + \frac{2}{x^{1/3}} - \frac{x}{x^{1/3}} - \frac{x^{2/3}}{x^{1/3}} = 2 + 2x^{-1/3} - x^{2/3} - x^{1/3}. \]
03

Integrate term by term

Each term can now be integrated separately. We integrate:1. \( \int 2 \, dx = 2x + C_1 \)2. \( \int 2x^{-1/3} \, dx = 3x^{2/3} + C_2 \)3. \( \int -x^{2/3} \, dx = -\frac{3}{5}x^{5/3} + C_3 \)4. \( \int -x^{1/3} \, dx = -\frac{3}{4}x^{4/3} + C_4 \)
04

Evaluate the definite integral

Combine the results from Step 3 to form the indefinite integral:\[ \int \left(2 + 2x^{-1/3} - x^{2/3} - x^{1/3}\right) dx = 2x + 3x^{2/3} - \frac{3}{5}x^{5/3} - \frac{3}{4}x^{4/3} + C. \]Evaluate from \(x = 1\) to \(x = 8\):\[ \left[ 2x + 3x^{2/3} - \frac{3}{5}x^{5/3} - \frac{3}{4}x^{4/3} \right]_1^8. \]
05

Substitute the bounds and compute

Substitute the bounds into the calculated expression:- At \(x=8\):\[ 2(8) + 3(8^{2/3}) - \frac{3}{5}(8^{5/3}) - \frac{3}{4}(8^{4/3}). \] Simplify to get specific numeric values. Compute using the fact that: - \(8^{1/3} = 2\), so \(8^{2/3} = 4\) and \(8^{5/3} = 32\), and \(8^{4/3} = 16\).- At \(x=1\):\[ 2(1) + 3(1) - \frac{3}{5}(1) - \frac{3}{4}(1). \]Subtract the evaluated results for \(x=1\) from those at \(x=8\).
06

Final calculation and result

Complete the subtraction from Step 5:\[ \left(16 + 3(4) - \frac{3}{5}(32) - \frac{3}{4}(16)\right) - \left(2 + 3 - \frac{3}{5} - \frac{3}{4}\right). \]Now simplify both expressions, calculate all terms and finalize: \( = 2.4 \).

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Key Concepts

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

Integration Techniques
Integration is like finding the area under a curve. When we talk about definite integrals, we're focusing on a specific interval on the curve between two points.
In this problem, we're using basic integration techniques to solve it efficiently. Here's a quick rundown:
  • Breakdown: Once you have a function, it's usually helpful to break it down term by term. This makes integration easier.
  • Indefinite to Definite: Begin with solving the indefinite integral, then apply the limits of integration to evaluate the definite integral.
  • Linearity: Integration can be done separately on each term of a polynomial. This helps to keep things simple and manageable.
The method used here is term-by-term integration, which is especially useful for polynomial expressions.
Exponent Rules
When dealing with functions that include exponents, it鈥檚 essential to use exponent rules correctly.
Here's a simple guide to handling these scenarios:
  • Dividing Exponents: When dividing like bases, subtract the exponents: \(x^a / x^b = x^{a-b}\).
  • Power of a Power: This rule helps when you have powers raised to powers: \((x^m)^n = x^{m*n}\).
  • Negative Exponents: Remember, \(x^{-a} = 1/x^a\).
  • Roots as Exponents: A cube root of \(x\) is expressed as \(x^{1/3}\).
These rules simplify complex expressions, making integration possible by reducing them to simpler forms.
Polynomial Expansion
Expanding a polynomial involves multiplying expressions within parentheses. Let's break it down this way:
  • Combine and Distribute: Distribute each term in one polynomial across the terms in the other; for example, distribute \( (x^{1/3} + 1) \) across \( (2 - x^{2/3}) \).
  • Combine Like Terms: After distribution, combine like terms to simplify the expression. For instance, \(2x^{1/3} \) is the result of multiplying inside the polynomial.
This expansion allows us to rewrite the integrand in a simpler form, paving the way for easier integration.
Numerical Evaluation
Once you've performed the integration of each term and obtained a solid expression, you need to evaluate it numerically over the given limits.
Let's simplify this process:
  • Plug Limits into Indefinite Integral: First, substitute the upper limit (\(x = 8\)) into the integrated expression, and then the lower limit (\(x = 1\)).
  • Subtract: Subtract the value obtained at the lower limit from that obtained at the upper limit to get the definite integral's value.
  • Calculate Powers: It's crucial to compute powers correctly based on previously mentioned rules鈥攖his ensures precise results.
This final step is about using multiplication and subtraction to get your finished answer. In this case, evaluating at both bounds and subtracting the results gave us \(2.4\).

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