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Magazine Chapter 11: Problem 19
Evaluate the definite integral. $$ \int_{1 / 2}^{2} \frac{d x}{\sqrt{2 x}(\sqrt{2 x}+9)} $$
Short Answer
Expert verified
\(\frac{1}{9}ln(\frac{2\sqrt{2}}{110})\)
Step by step solution
01
Simplify the integrand
First, recognize that the integrand can be simplified by a substitution. Let鈥檚 set the substitution variable and simplify the integral. Let \(u = \sqrt{2x}\). Then \(u^2 = 2x\) and \(dx = \frac{u du}{2}\).
02
Change the limits of integration
When \(x = \frac{1}{2}\), \(u = \sqrt{2 \frac{1}{2}} = 1\). When \(x = 2\), \(u = \sqrt{2*2} = 2\sqrt{2}\). So the new limits of integration are from \(1\) to \(2\sqrt{2}\).
03
Substitute and integrate
Substitute the new variables into the integral: \( \int_{1}^{2\sqrt{2}} \frac{du}{u(u+9)} \). This can be decomposed into partial fractions: \( \frac{1}{u(u+9)} = \frac{A}{u} + \frac{B}{u+9} \). Solving this gives \(A = \frac{1}{9}\) and \(B = \frac{-1}{9}\).
04
Integrate the partial fractions
Rewrite the integral as \(\frac{1}{9}\int_{1}^{2\sqrt{2}} \frac{1}{u} du - \frac{1}{9}\int_{1}^{2\sqrt{2}} \frac{1}{u+9} du \). These integrals can be simplified to logarithms: \(\frac{1}{9}(ln|u| - ln|u+9|) \Bigg|_1^{2\sqrt{2}}\).
05
Simplify the result
Evaluate the integral: \[ \frac{1}{9}(ln(2\sqrt{2}) - ln(11)) - (ln(1) - ln(10))\]. Simplify the result to get the answer: \[ \frac{1}{9}(ln(2\sqrt{2}) - ln(11) - ln(10)) \].
06
Final simplification
Combine the logarithms to simplify the expression: \[ \frac{1}{9}ln\left(\frac{2\sqrt{2}}{110}\right) \]. This is the final result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
integration techniques
When evaluating a definite integral, there are various techniques you can use. The right method depends on the form of the integrand. Here are some basic techniques:
- **Substitution Method**: Used when an integral contains a function and its derivative. It simplifies the integral by transforming it into a basic form.
- **Partial Fractions**: This helps when working with rational functions (ratios of polynomials). Decomposing these functions into simpler fractions makes them easier to integrate.
- **Integration by Parts**: It is helpful when the integrand is a product of two functions. This method uses the formula: \ \ \ \( \int u dv = uv - \int v du \).
substitution method
The substitution method involves replacing a part of the integrand with a new variable. This helps simplify the expression in the integral. Let's look step-by-step:
- Step 1: Choose a substitution: For the integral \ \( \int \frac{d x}{\sqrt{2 x}(\sqrt{2 x}+9)} \) we chose \( u = \sqrt{2x} \).
- Step 2: Find the differential: From \( u = \sqrt{2x} \), we get \( u^2 = 2x \implies dx = \frac{u du}{2} \).
- Step 3: Change limits: When \( x = \frac{1}{2} \), \(u = 1\). When \( x = 2 \), \(u = 2\sqrt{2}\).
partial fractions
Partial fractions are useful when dealing with integrals of rational functions. These functions can be broken down into simpler, more manageable pieces. Here's how we do this:
- Step 1: Set Up Partial Fractions: Decompose \( \frac{1}{u(u+9)} \). We assume \ \( \frac{1}{u(u+9)} = \frac{A}{u} + \frac{B}{u+9} \).
- Step 2: Solve for Constants: Multiply both sides by \( u(u+9) \) and set up an equation to find \(A\) and \(B\). We get \( A(u+9) + Bu = 1 \). Solving, we find \( A = \frac{1}{9} \) and \( B = -\frac{1}{9} \).
- Step 3: Rewrite Integral: Use these constants to rewrite the integral: \( \frac{1}{9}\int \frac{1}{u} du - \frac{1}{9}\int \frac{1}{u+9} du \).
