Chapter 6: Problem 45
Find the indefinite integrals. $$\int \frac{8}{\sqrt{x}} d x$$
Short Answer
Expert verified
The indefinite integral is \( 16\sqrt{x} + C \).
Step by step solution
01
Understand the Integral
We need to find the indefinite integral of the function \( \frac{8}{\sqrt{x}} \). This function can be rewritten in a form that is easier to integrate by expressing the square root in terms of exponents.
02
Rewrite the Integrand
The function \( \frac{8}{\sqrt{x}} \) can be rewritten as \( 8x^{-\frac{1}{2}} \). This is because \( x^{-\frac{1}{2}} \) is equivalent to \( \frac{1}{\sqrt{x}} \).
03
Apply the Power Rule for Integration
To integrate \( 8x^{-\frac{1}{2}} \), we use the power rule which states that \( \int x^n \: dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \). For \( 8x^{-\frac{1}{2}} \), \( n = -\frac{1}{2} \).
04
Perform the Integration
Applying the power rule: \[\int 8x^{-\frac{1}{2}} \: dx = 8 \cdot \frac{x^{(-\frac{1}{2})+1}}{(-\frac{1}{2})+1} + C = 8 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C\] Simplifying further gives us:\[= 8 \cdot 2x^{\frac{1}{2}} + C = 16\sqrt{x} + C\]
05
State the Indefinite Integral
Therefore, the indefinite integral of \( \frac{8}{\sqrt{x}} \) is \( 16\sqrt{x} + C \), 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.
Power Rule for Integration
The **Power Rule for Integration** is a fundamental concept in calculus that allows us to find antiderivatives of functions that are powers of variables. To use this rule, we consider functions of the form \( x^n \), where \( n \) is a constant other than -1.
Here's how it works:- If a function is \( x^n \), its integral is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), with \( C \) the constant of integration.- When \( n = -1 \), this form doesn't work because it leads to a division by zero. For \( n = -1 \), the integral is the natural logarithm \( \ln|x| + C \). This rule is useful for integration problems like \( 8x^{-\frac{1}{2}} \). We increase the exponent by 1 and then divide by the new exponent. With practice, this technique becomes quite intuitive, helping you solve a wide range of integrals quickly.
Here's how it works:- If a function is \( x^n \), its integral is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), with \( C \) the constant of integration.- When \( n = -1 \), this form doesn't work because it leads to a division by zero. For \( n = -1 \), the integral is the natural logarithm \( \ln|x| + C \). This rule is useful for integration problems like \( 8x^{-\frac{1}{2}} \). We increase the exponent by 1 and then divide by the new exponent. With practice, this technique becomes quite intuitive, helping you solve a wide range of integrals quickly.
Integration Techniques
Integration techniques are strategies used to solve integrals, and they are crucial when handling more complex functions. Here are a few key techniques:- **Rewriting expressions**: Transform expressions into simpler forms. For instance, changing \( \frac{8}{\sqrt{x}} \) to \( 8x^{-\frac{1}{2}} \) helps you apply the Power Rule more easily.- **Substitution**: Used when a function is a composite of another. It simplifies integrals that aren’t directly solvable via standard rules.- **Partial Fraction Decomposition**: Works well with rational functions, it breaks them down into simpler fractions.
Mastering these techniques enables you to tackle a variety of integration problems. In our example with \( 8x^{-\frac{1}{2}} \), using the trick of rewriting helped make the integration process simpler and more direct.
Mastering these techniques enables you to tackle a variety of integration problems. In our example with \( 8x^{-\frac{1}{2}} \), using the trick of rewriting helped make the integration process simpler and more direct.
Exponents and Roots
Understanding **Exponents and Roots** aids greatly in calculus, especially when dealing with integrals. Exponents represent how many times to multiply a number by itself, while roots such as square roots are the reverse process.
Important points on this topic include: - **Changing Roots to Exponents**: Recognizing that \( \sqrt{x} \) is equivalent to \( x^{\frac{1}{2}} \) can simplify complex problems.- **Negative Exponents**: These signify reciprocals (i.e., \( x^{-n} = \frac{1}{x^n} \)), and can be particularly helpful in algebraic simplification.For example, \( \frac{8}{\sqrt{x}} \) is rewritten as \( 8x^{-\frac{1}{2}} \) to make the integral easily solvable using the Power Rule. Learning these relationships enhances your problem-solving toolkit and supports better understanding of mathematical expressions.
Important points on this topic include: - **Changing Roots to Exponents**: Recognizing that \( \sqrt{x} \) is equivalent to \( x^{\frac{1}{2}} \) can simplify complex problems.- **Negative Exponents**: These signify reciprocals (i.e., \( x^{-n} = \frac{1}{x^n} \)), and can be particularly helpful in algebraic simplification.For example, \( \frac{8}{\sqrt{x}} \) is rewritten as \( 8x^{-\frac{1}{2}} \) to make the integral easily solvable using the Power Rule. Learning these relationships enhances your problem-solving toolkit and supports better understanding of mathematical expressions.