Chapter 28: Problem 13
Integrate each of the given functions. $$\int_{0}^{1} \frac{2 t+4}{3 t^{2}+5 t+2} d t$$
Short Answer
Expert verified
The integral evaluates to \(\frac{8}{3} \ln\left( \frac{5}{2} \right) - \ln(4)\).
Step by step solution
01
Recognize the Partial Fraction Decomposition
The function inside the integral is a rational function. We use partial fraction decomposition to simplify it. The denominator is given by \(3t^2 + 5t + 2\). We need to factor this quadratic.
02
Factor the Quadratic
Factor the quadratic \(3t^2 + 5t + 2\). It can be factored as \((3t + 2)(t + 1)\). So, we rewrite the fraction \(\frac{2t+4}{3t^2 + 5t + 2}\) using these factors.
03
Set Up the Partial Fractions
Express \(\frac{2t+4}{(3t + 2)(t + 1)}\) as the sum of two fractions: \(\frac{A}{3t+2} + \frac{B}{t+1}\), where \(A\) and \(B\) are constants to find.
04
Solve for 'A' and 'B'
Multiply through by the denominator \((3t+2)(t+1)\) to clear fractions: \[2t + 4 = A(t + 1) + B(3t + 2)\]Expand and combine like terms:\[2t + 4 = (A + 3B)t + (A + 2B)\]Equate coefficients to solve for \(A\) and \(B\):1. \(A + 3B = 2\)2. \(A + 2B = 4\)Solving these equations, \(B = -2\) and \(A = 8\).
05
Substitute Back and Integrate
Substitute \(A = 8\) and \(B = -2\) back into the partial fractions:\[\int_0^1 \left( \frac{8}{3t+2} - \frac{2}{t+1} \right) dt\]Integrate each term separately:1. \(\int \frac{8}{3t+2} dt = \frac{8}{3} \ln|3t+2|\)2. \(\int -\frac{2}{t+1} dt = -2 \ln|t+1|\)Thus, the integral becomes:\[\frac{8}{3} \ln|3t+2| - 2 \ln|t+1|\]_0^1.
06
Evaluate the Integral from 0 to 1
Substitute the limits into the integrated function:\[\left( \frac{8}{3} \ln(5) - 2 \ln(2) \right) - \left( \frac{8}{3} \ln(2) - 2 \ln(1) \right)\]Simplify using \(\ln(1) = 0\):\[\frac{8}{3} \ln(5) - 2 \ln(2) - \left( \frac{8}{3} \ln(2) \right) = \frac{8}{3} (\ln(5) - \ln(2)) - 2 \ln(2)\]Recognize \(\ln(5) - \ln(2) = \ln(\frac{5}{2})\) and simplify further:\[\frac{8}{3} \ln\left( \frac{5}{2} \right) - 2 \ln(2)\].
07
Final Simplification
Simplify the expression using logarithmic properties:\[\frac{8}{3} \ln\left( \frac{5}{2} \right) - 2 \ln(2) = \frac{8}{3} \ln\left( \frac{5}{2} \right) - \ln(4)\].This gives the final evaluated result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
When dealing with complex rational functions, **partial fraction decomposition** is your go-to tool to simplify integration. The essence of this technique is to break down a complicated fraction into simpler, separate parts, making it easier to handle. Consider a function with a polynomial in the denominator: if it can be factored into linear factors like
- 3t + 2
- t + 1
- \( \frac{A}{3t+2} \)
- \( \frac{B}{t+1} \)
Definite Integral
A **definite integral** evaluates the area under a curve over a specified interval. For example, evaluating from 0 to 1 of our decomposed fraction ensures we measure precisely how much 'space' exists between the curve of the function and the x-axis from 0 to 1.
To properly evaluate a definite integral, follow these steps:
- Integrate the function to find the antiderivative.
- Substitute the upper limit, then the lower limit into the antiderivative.
- Subtract the value at the lower limit from the value at the upper limit.
Integration Techniques
While integrating complex functions, employing various **integration techniques** can greatly simplify the process. One such method is reducing rational functions using partial fraction decomposition, as discussed.
Let's highlight some common techniques:
- Substitution: Replacing variables to make integrals easier or possible. Very handy in trigonometric integrals.
- Integration by Parts: Based on the product rule for differentiation, this method is useful for integrating products of functions.
- Recognizing Patterns: Many integrals fit common patterns. Recognizing these can save significant time.
Logarithmic Integration
In our exercise, the final integral involved two terms, each of which required **logarithmic integration**. This integration technique is primarily used when you have integrals of the form:\[ \int \frac{c}{at+b} \, dt \]The solution here would typically be:\[ \frac{c}{a} \ln|at+b| + C \]And that’s exactly what we did. For both fractions after decomposition, each term resulted in a logarithmic expression:
- \( \frac{8}{3} \ln|3t+2| \)
- \(-2 \ln|t+1| \)