Question

For the following exercises, evaluate by any method. $$ \int_{5}^{10} \frac{d t}{t}-\int_{5 x}^{10 x} \frac{d t}{t} $$

Short answer

The evaluated expression is 0.

Step-by-step solution

Simplify the Expression

The given expression is the difference between two integrals: \[ \int_{5}^{10} \frac{d t}{t} - \int_{5x}^{10x} \frac{d t}{t}. \]Apply the property of integrals which states that for any two functions \(f(t)\) and suitable limits \(a, b, c\): \[\int_{a}^{b} f(t) \, dt + \int_{b}^{c} f(t) \, dt = \int_{a}^{c} f(t) \, dt \]to rewrite the second integral: \[\int_{5x}^{10x} \frac{d t}{t} = -\int_{10x}^{5x} \frac{d t}{t}.\]Thus, our expression becomes: \[\int_{5}^{10} \frac{d t}{t} + \int_{10x}^{5x} \frac{d t}{t}. \]

Evaluate the First Integral

Evaluate the integral \(\int_{5}^{10} \frac{d t}{t}\). The antiderivative of \(\frac{1}{t}\) is \(\ln|t|\). So, the evaluation gives: \[\left[ \ln|t| \right]_{5}^{10} = \ln|10| - \ln|5| = \ln\left(\frac{10}{5}\right) = \ln(2). \]

Evaluate the Second Integral with Change of Limits

Evaluate \(\int_{10x}^{5x} \frac{d t}{t}\): The antiderivative is also \(\ln|t|\). So, \[\left[ \ln|t| \right]_{10x}^{5x} = \ln|5x| - \ln|10x| = \ln\left(\frac{5x}{10x}\right) = \ln\left(\frac{1}{2}\right). \]Note that the change from \(10x\) to \(5x\) flipped the sign of the integral.

Combine the Results

Now combine the results from both integrals: \[\ln(2) + \ln\left(\frac{1}{2}\right). \]Using the property of logarithms \(\ln(a) + \ln(b) = \ln(ab)\), combine: \[\ln(2) + \ln\left(\frac{1}{2}\right) = \ln(2 \times \frac{1}{2}) = \ln(1) = 0. \]

Conclusion

The evaluated value of the given expression \(\int_{5}^{10} \frac{d t}{t} - \int_{5x}^{10x} \frac{d t}{t}\) is \(0\). This result illustrates that the two integrals cancel each other exactly regardless of \(x\).

Key concepts

Integral Properties

Integral properties are essential tools when working with definite integrals. They help simplify complex expressions by using known rules about how integrals behave. One key property is the linearity of the integral, which involves the sum of integrals:

• For two functions, \(f(t)\) and \(g(t)\), the rule \(\int (f(t) + g(t)) \, dt = \int f(t) \, dt + \int g(t) \, dt\).

Another important property is the ability to reverse the limits of integration, which changes the sign of the integral:

• \(\int_{a}^{b} f(t) \, dt = -\int_{b}^{a} f(t) \, dt\).

Using these properties can significantly simplify the process of evaluating integrals by allowing you to adjust the integral expressions in ways that make them easier to compute.

Logarithm Rules

Logarithm rules are powerful tools for simplifying expressions involving logs, especially when dealing with integrals where the functions take the form of \(\frac{1}{t}\). The main rules to keep in mind are:

• The product rule: \( \ln(a) + \ln(b) = \ln(ab) \).
• The quotient rule: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \).
• The power rule: \( \ln(a^b) = b \cdot \ln(a) \).

In the context of definite integrals, especially those resulting in logarithmic expressions, these rules help combine or simplify the final results, leading to easier computations.

Antiderivative

The antiderivative represents the most general form of a function whose derivative is the given function. In the context of integrals, finding the antiderivative is a crucial step in solving definite integrals. For the function \(\frac{1}{t}\), the antiderivative is \(\ln|t| + C\), where \(C\) is the arbitrary constant of integration.

When evaluating definite integrals, however, the constant \(C\) cancels out in the solution, so it's not explicitly required. Understanding the concept of the antiderivative is key when applying the fundamental theorem of calculus to compute the value of a definite integral from its limits.

Evaluating Definite Integrals

Evaluating definite integrals involves finding the difference between the values of the antiderivative at two points. This process is based on the fundamental theorem of calculus. The steps generally include:

• Finding the antiderivative of the integrand.
• Substituting the upper and lower bounds into the antiderivative function and calculating the difference.

For the given integrals, this boils down to computing expressions like \([\ln|t|]_{a}^{b} = \ln|b| - \ln|a|\). This simplifies further using logarithm rules.

The remarkable part of definite integrals is how they can cancel each other out or add up, as seen in the original exercise, which resulted in zero due to the subtraction and addition properties of the integrals used.