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Magazine Chapter 11: Problem 20
Evaluate the definite integral. $$ \int_{0}^{1} \frac{9 d x}{8 x^{3}+1} $$
Short Answer
Expert verified
36(ln(3/1)) = 36 ln(3)
Step by step solution
01
Identify substitution
The integral given is \( \int_{0}^{1} \frac{9 \, dx}{8 x^{3} + 1} \). Notice that the denominator can be factored using the identity for the sum of cubes \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). To simplify this, let鈥檚 find a substitution. We can set \( u = 2x \), which implies \( du = 2dx \), so \( dx = \frac{du}{2} \).
02
Apply the substitution
Substitute \( x = \frac{u}{2} \) into the integral: \( x^3 = \left( \frac{u}{2} \right)^3 = \frac{u^3}{8} \). The integral becomes: \[ \int_{0}^{2} \frac{9 \cdot \frac{du}{2}}{8 \cdot \left( \frac{u}{2} \right)^3 + 1} = \int_{0}^{2} \frac{9 \cdot \frac{du}{2}}{\frac{u^3}{2} + 1} = \int_{0}^{2} \frac{9 \cdot \frac{du}{2}}{u^3/8 + 1} \]
03
Simplify the integral
Simplify the resulting integral: \[ \int_{0}^{2} \frac{9/2 \, du}{u^3/8 + 1} \] = \[ \int_{0}^{2} \frac{9/2 \, du}{(u^3 + 8) / 8} \] = \[ \int_{0}^{2} \frac{9/2 \, du \cdot 8}{u^3 + 8} \] = \[ \int_{0}^{2} \frac{36 \, du}{u^3 + 8} \]
04
Factor the denominator
Factor the denominator \( u^3 + 8 \) using the sum of cubes: \( u^3 + 8 = (u + 2)(u^2 - 2u + 4) \). The integral becomes: \[ \int_{0}^{2} \frac{36 \, du}{(u + 2)(u^2 - 2u + 4)} \]
05
Perform partial fractions decomposition
Use partial fractions to decompose the integrand: \[ \frac{36}{(u + 2)(u^2 - 2u + 4)} = \frac{A}{u + 2} + \frac{Bu + C}{u^2 - 2u + 4} \]. Solve for constants A, B, and C by equating coefficients of corresponding powers of u.
06
Integrate each part
Once the partial fractions decomposition has been determined: \[ \int_{0}^{2} \frac{A}{u + 2} \, du + \int_{0}^{2} \frac{Bu + C}{u^2 - 2u + 4} \, du \]. Integrate each part separately using standard integral formulas.
07
Evaluate the definite integral
After integrating, substitute the limits back into the solutions to evaluate the definite integral from 0 to 2.
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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 technique used to simplify the evaluation of integrals. By replacing a part of the integrand with a new variable, we can transform the integral into a simpler form. In this exercise, we identified the substitution as \( u = 2x \), which allows us to rewrite the variable and its differential. When applying the substitution, remember to adjust the differential accordingly. In this case, \( du = 2dx \) led us to \( dx = \frac{du}{2} \). This simplifies the complex integral into a more manageable form for further steps.
Partial Fractions Decomposition
Partial fractions decomposition is often used to break down a complex rational function into simpler fractions. To decompose the function, we express it as a sum of simpler fractions:
- First, factor the denominator, as done in the solution where it was factored into \( u^3 + 8 = (u + 2)(u^2 - 2u + 4) \)
Sum of Cubes Formula
The sum of cubes formula is useful to factor expressions like \( a^3 + b^3 \). This formula states that:
- \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)
Integration Techniques
There are various techniques used to solve integrals; understanding which to use when simplifies the process significantly:
- Substitution: Transforms integrals into easier forms, as seen when \( u = 2x \) simplified our complex integral.
- Partial Fractions: Conducted after factoring simplifies rational functions, useful when the integrand is a ratio of polynomials.
- Recognizing Special Forms: Utilizing algebraic identities like the sum of cubes formula simplifies polynomial expressions.
