Chapter 6: Problem 61
Evaluate the following definite integrals. Use Theorem 10 to express your answer in terms of logarithms. \(\int_{-2}^{2} \frac{d t}{t^{2}-9}\)
Short Answer
Expert verified
Question: Evaluate the integral \(\int_{-2}^{2} \frac{dt}{t^2-9}\)
Answer: \(\frac{1}{6}ln|\frac{1}{5}|\)
Step by step solution
01
Apply partial fractions decomposition
We need to express the rational function \(\frac{1}{t^2-9}\) as the sum of two simpler fractions with linear factors in the denominator. We decompose the fraction into partial fractions:
\(\frac{1}{t^2-9} = \frac{A}{t-3} + \frac{B}{t+3}\)
We must now solve for the constants A and B.
02
Solve for A and B
To solve for A and B, we first clear the denominators by multiplying both sides of the equation by \((t-3)(t+3)\):
\(1 = A (t+3) + B(t-3)\)
Now we can find the values of A and B by solving a system of equations. One approach is to use substitution.
Let \(t = 3\), then the equation becomes:
\(1 = A(6)\), which leads to \(A=\frac{1}{6}\)
Let \(t = -3\), then the equation becomes:
\(1 = -B(6)\), which leads to \(B=-\frac{1}{6}\)
Now that we have the values of A and B, we can rewrite our integrand:
\(\frac{1}{t^2-9} = \frac{\frac{1}{6}}{t-3} - \frac{\frac{1}{6}}{t+3}\)
03
Integrate term by term
Now that we have our integrand in terms of partial fractions, we can evaluate the definite integral term by term:
\(\int_{-2}^{2} \frac{1}{t^2-9} dt = \int_{-2}^{2} \left(\frac{\frac{1}{6}}{t-3}-\frac{\frac{1}{6}}{t+3}\right) dt\)
Using the properties of logarithms, we get:
\(= \frac{1}{6}[\int_{-2}^{2} \frac{1}{t-3} dt - \int_{-2}^{2} \frac{1}{t+3} dt]\)
Now, we integrate each term:
\(= \frac{1}{6}[ln|t-3| - ln|t+3|]\Big|_{-2}^{2}\)
04
Evaluate the integral
Finally, we need to evaluate the definite integral. Plug in the limits of integration and simplify:
\(= \frac{1}{6}[ln|\frac{t-3}{t+3}|]\Big|_{-2}^{2}\)
\(= \frac{1}{6}[ln|\frac{2-3}{2+3}| - ln|\frac{-2-3}{-2+3}|]\)
\(= \frac{1}{6}[ln|\frac{-1}{5}| - ln|-1|]\)
\(= \frac{1}{6}\left[ln|\frac{1}{5}| \right]\)
The final answer is:
\(= \frac{1}{6}ln|\frac{1}{5}|\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
When you encounter a complex fraction that you need to integrate, partial fraction decomposition can be your go-to tool. This technique helps break down a complicated rational function into simpler fractions, making integration much more straightforward.
In the given problem, you start with the function \(\frac{1}{t^2-9}\). Notice that the denominator \(t^2-9\) can be factored into \((t-3)(t+3)\). This factorization allows you to express the function as a sum of two simpler fractions with linear denominators:
\[ \frac{1}{t^2-9} = \frac{A}{t-3} + \frac{B}{t+3} \]
To find the values of \(A\) and \(B\), you clear the denominator by multiplying through by \((t-3)(t+3)\), resulting in:
\[ 1 = A(t+3) + B(t-3) \]
By carefully choosing values for \(t\) (often the roots of the denominator), you solve this equation for \(A\) and \(B\). By letting \(t\) equal the roots \(3\) and \(-3\), you isolate each term, leading to \(A=\frac{1}{6}\) and \(B=-\frac{1}{6}\). This simplifies the integrand into terms that are easier to handle individually through integration.
In the given problem, you start with the function \(\frac{1}{t^2-9}\). Notice that the denominator \(t^2-9\) can be factored into \((t-3)(t+3)\). This factorization allows you to express the function as a sum of two simpler fractions with linear denominators:
\[ \frac{1}{t^2-9} = \frac{A}{t-3} + \frac{B}{t+3} \]
To find the values of \(A\) and \(B\), you clear the denominator by multiplying through by \((t-3)(t+3)\), resulting in:
\[ 1 = A(t+3) + B(t-3) \]
By carefully choosing values for \(t\) (often the roots of the denominator), you solve this equation for \(A\) and \(B\). By letting \(t\) equal the roots \(3\) and \(-3\), you isolate each term, leading to \(A=\frac{1}{6}\) and \(B=-\frac{1}{6}\). This simplifies the integrand into terms that are easier to handle individually through integration.
Logarithmic Integration
After breaking the function into partial fractions, you integrate each term separately. This process often reveals logarithmic integration, a method where you integrate functions that have fractions as their integrands, resulting in logarithmic expressions.
For the given fractions \(\frac{1}{6} \frac{1}{t-3}\) and \(-\frac{1}{6} \frac{1}{t+3}\), the integrals become:
\[ \int \frac{1}{t-3} dt \,=\, \ln|t-3| \text{ and } \int \frac{1}{t+3} dt \,=\, \ln|t+3| \]
Always remember to include absolute value signs in the logarithm because they ensure the function's validity over its domain.
Once integrated, apply the limits of integration to compute the definite integral. The result becomes a difference of natural logarithms due to the subtraction sign between the two integrals. Thus,
\[ \frac{1}{6}[\ln|t-3| - \ln|t+3|]\Bigg|_{-2}^{2} \]
becomes your expression to evaluate, based on identifying the limits \(-2\) to \(2\) into your logarithms.
For the given fractions \(\frac{1}{6} \frac{1}{t-3}\) and \(-\frac{1}{6} \frac{1}{t+3}\), the integrals become:
\[ \int \frac{1}{t-3} dt \,=\, \ln|t-3| \text{ and } \int \frac{1}{t+3} dt \,=\, \ln|t+3| \]
Always remember to include absolute value signs in the logarithm because they ensure the function's validity over its domain.
Once integrated, apply the limits of integration to compute the definite integral. The result becomes a difference of natural logarithms due to the subtraction sign between the two integrals. Thus,
\[ \frac{1}{6}[\ln|t-3| - \ln|t+3|]\Bigg|_{-2}^{2} \]
becomes your expression to evaluate, based on identifying the limits \(-2\) to \(2\) into your logarithms.
Theorem of Calculus
The theorem of calculus, known as the Fundamental Theorem of Calculus, bridges the concept of antiderivatives with the calculation of definite integrals. The theorem states that if a function \(f\) is continuous on the closed interval \([a, b]\), then the integral of \(f\) from \(a\) to \(b\) is found by evaluating an antiderivative at these endpoints.
Making use of this theorem in our exercise is crucial. We perform this by substituting the endpoints into the antiderivative obtained through partial fractions and logarithmic integration.
For instance, compute \(\ln|\frac{t-3}{t+3}|\) at the bounds \(2\) and \(-2\),
leading to:
\[ \frac{1}{6}[\ln|\frac{2-3}{2+3}| - \ln|\frac{-2-3}{-2+3}|] \]
Simplifying these terms gives a finite logarithmic expression. This allows you to write the final answer as:
\[ \frac{1}{6} \ln|\frac{1}{5}| \]
Usage of the theorem of calculus transforms complex chasing of areas under curves into a straightforward application of numeric evaluations, demonstrating its powerful utility in calculus.
Making use of this theorem in our exercise is crucial. We perform this by substituting the endpoints into the antiderivative obtained through partial fractions and logarithmic integration.
For instance, compute \(\ln|\frac{t-3}{t+3}|\) at the bounds \(2\) and \(-2\),
leading to:
\[ \frac{1}{6}[\ln|\frac{2-3}{2+3}| - \ln|\frac{-2-3}{-2+3}|] \]
Simplifying these terms gives a finite logarithmic expression. This allows you to write the final answer as:
\[ \frac{1}{6} \ln|\frac{1}{5}| \]
Usage of the theorem of calculus transforms complex chasing of areas under curves into a straightforward application of numeric evaluations, demonstrating its powerful utility in calculus.
