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Magazine Chapter 25: Problem 32
Integrate each of the given expressions. $$\int \frac{8 d V}{(0.3+2 V)^{3}}$$
Short Answer
Expert verified
\( \int \frac{8 dV}{(0.3+2V)^3} = -\frac{2}{(0.3+2V)^2} + C \).
Step by step solution
01
Identify the Integral Type
The integral given is \( \int \frac{8 \, dV}{(0.3 + 2V)^3} \). This forms a rational function where the numerator is a constant and the denominator is a power of a linear term.
02
Use Substitution Method
Let's use substitution to simplify the integration. Set \( u = 0.3 + 2V \). Then, the differential \( du = 2 \, dV \), or \( dV = \frac{1}{2} \, du \). The integral becomes \( \int \frac{8}{u^3} \times \frac{1}{2} \, du = 4 \int u^{-3} \, du \).
03
Integrate the Simplified Expression
Now integrate the new expression: \( 4 \int u^{-3} \, du \). The antiderivative of \( u^{-3} \) is \( \frac{u^{-2}}{-2} = -\frac{1}{2u^2} \), so the integral becomes \( 4 \left( -\frac{1}{2u^2} \right) = -\frac{2}{u^2} \).
04
Substitute Back to Original Variable
Substitute back \( u = 0.3 + 2V \) into the integral. The expression becomes \( -\frac{2}{(0.3 + 2V)^2} \).
05
Add the Constant of Integration
Don't forget the constant of integration when indefinite integrals are involved. Hence, the final answer is \( -\frac{2}{(0.3 + 2V)^2} + C \), where \( C \) is the constant of integration.
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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 clever technique used to simplify integrals. It transforms a complicated problem into an easier one. This technique is especially handy when dealing with integrals that involve composite functions. Here's how it works:
- Identify a part of the integral that can be replaced with a single variable, say \( u \). This substitution should simplify the whole expression.
- Compute the differential, \( du \), in terms of the original variable using the derivative of your substitution.
- Replace the original variable and its differential with \( u \) and \( du \) in the integral.
- Integrate in terms of \( u \), which is often simpler than dealing with the original expression.
- Finally, substitute back the original variable to express the result in terms of the original variable.
Rational Functions
Rational Functions play a crucial role in calculus. They are quotients of two polynomials and often appear in integration problems.
In the exercise, we dealt with the function \( \frac{8}{(0.3+2V)^3} \). This is a perfect example of a rational function, where the numerator is constant, and the denominator is a power of a linear term.
In the exercise, we dealt with the function \( \frac{8}{(0.3+2V)^3} \). This is a perfect example of a rational function, where the numerator is constant, and the denominator is a power of a linear term.
- Integrating rational functions can either be straightforward or require some manipulations like partial fraction decomposition or substitution.
- The step-by-step integration of rational functions might first involve expressing them in a more convenient form.
- In cases where the denominator is a single term raised to a power, substitution is often employed to reduce complexity.
Indefinite Integrals
Indefinite Integrals are the opposite of derivatives. Unlike definite integrals that provide a numerical result, indefinite integrals represent families of functions, as they include a constant of integration.
- An indefinite integral has the form \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative of \( f(x) \).
- When evaluating indefinite integrals, the constant of integration, denoted as \( C \), must always be added as it accounts for all possible vertical shifts of the antiderivative.
- This constant arises because differentiation of any constant results in zero, making the original function undetermined by a set amount when you integrate.
- In the provided exercise, after substituting and integrating, we achieve an antiderivative of the function, ultimately leading to \( -\frac{2}{(0.3 + 2V)^2} + C \). Including the constant \( C \) is vital as it completes the solution by acknowledging all potential solutions of the integral.
