Problem 34
Question
Evaluate the given definite integral by finding an antiderivative of the integrand and applying Theorem \(3 .\) $$ \int_{2}^{3}\left(6 / x^{2}\right) d x $$
Step-by-Step Solution
Verified Answer
The value of the integral is 1.
1Step 1: Identify the Integrand
The integrand in the given definite integral is \( \frac{6}{x^2} \). To simplify this, you can rewrite it in a form that is easier to integrate: \( 6x^{-2} \).
2Step 2: Find the Antiderivative
To find the antiderivative of \( 6x^{-2} \), use the power rule for integration. The power rule states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), for \( n eq -1 \). Apply the power rule: \( \int 6x^{-2} \, dx = 6 \cdot \frac{x^{-1}}{-1} = -6x^{-1} + C = -\frac{6}{x} + C \).
3Step 3: Apply the Limits of Integration
Now that you have the antiderivative \( -\frac{6}{x} \), apply the limits from \(2\) to \(3\). Using the Fundamental Theorem of Calculus, we calculate: \[-\frac{6}{3} - \left( -\frac{6}{2} \right)\].
4Step 4: Simplify the Expression
Perform the arithmetic: \[-\frac{6}{3} = -2\] and \(-\frac{6}{2} = -3\). The calculation becomes \(-2 + 3\).
5Step 5: Final Calculation
Add the results from the simplification: \(-2 + 3 = 1\). This value is the evaluated result of the definite integral.
Key Concepts
AntiderivativePower RuleFundamental Theorem of Calculus
Antiderivative
The antiderivative, often referred to as the indefinite integral, is a function that reverses the process of differentiation. In simple terms, it's the opposite of finding a derivative. When you integrate a function, you're looking for a function whose derivative is the original function you started with.
For example, if you have a function like \( f(x) = x^2 \), its derivative is \( f'(x) = 2x \). Conversely, when integrating \( 2x \), you end up with \( x^2 + C \), where \( C \) represents the constant of integration. This concept is critical when solving integrals because it helps you uncover the original function from its rate of change.
In the case of the exercise above, converting \( \frac{6}{x^2} \) to \( 6x^{-2} \) and applying the antiderivative gives \( -\frac{6}{x} + C \). Solving definite integrals often begins with finding the antiderivative like this.
For example, if you have a function like \( f(x) = x^2 \), its derivative is \( f'(x) = 2x \). Conversely, when integrating \( 2x \), you end up with \( x^2 + C \), where \( C \) represents the constant of integration. This concept is critical when solving integrals because it helps you uncover the original function from its rate of change.
In the case of the exercise above, converting \( \frac{6}{x^2} \) to \( 6x^{-2} \) and applying the antiderivative gives \( -\frac{6}{x} + C \). Solving definite integrals often begins with finding the antiderivative like this.
Power Rule
The power rule for integration is a handy formula that simplifies the integration process when dealing with functions that are powers of \( x \). It's the opposite process of the power rule for differentiation.
The rule states: \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]with the condition that \( n eq -1 \). This restriction exists because dividing by zero is undefined, as would be the case with \( n = -1 \).
To apply this rule, increase the exponent \( n \) by one, divide by the new exponent, and don't forget to add the constant \( C \), acknowledging there can be multiple antiderivatives.
The rule states: \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]with the condition that \( n eq -1 \). This restriction exists because dividing by zero is undefined, as would be the case with \( n = -1 \).
To apply this rule, increase the exponent \( n \) by one, divide by the new exponent, and don't forget to add the constant \( C \), acknowledging there can be multiple antiderivatives.
- Example: To find the antiderivative of \( 6x^{-2} \), apply the power rule: add 1 to \( -2 \) to get \( -1 \), divide by \( -1 \), resulting in \(-\frac{6}{x} + C \).
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) connects the concept of derivatives with integrals, providing a bridge between these two core aspects of calculus. It comes in two parts, but the aspect relevant to definite integrals is the second part.
The theorem states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then:
\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]
In essence, it says that to compute a definite integral, evaluate the antiderivative at the upper limit and subtract the evaluation at the lower limit. This evaluation process converts the antiderivative into actual work like calculating distances or areas between curves.
Applying this principle in the exercise, you have the antiderivative \(-\frac{6}{x} \). Evaluating it from \( 2 \) to \( 3 \) yields the operations \(-\frac{6}{3} - \left(-\frac{6}{2}\right)\), simplifying to \(-2 + 3 \) and resulting in \( 1 \). Thus, the definite integral evaluates to \( 1 \).
The theorem states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then:
\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]
In essence, it says that to compute a definite integral, evaluate the antiderivative at the upper limit and subtract the evaluation at the lower limit. This evaluation process converts the antiderivative into actual work like calculating distances or areas between curves.
Applying this principle in the exercise, you have the antiderivative \(-\frac{6}{x} \). Evaluating it from \( 2 \) to \( 3 \) yields the operations \(-\frac{6}{3} - \left(-\frac{6}{2}\right)\), simplifying to \(-2 + 3 \) and resulting in \( 1 \). Thus, the definite integral evaluates to \( 1 \).
- This theorem allows you to compute integrals precisely, offering a practical tool for solving real-world problems efficiently.
Other exercises in this chapter
Problem 34
Use the method of substitution to evaluate the definite integrals. $$ \int_{4}^{9} t^{-3 / 2} \sin (\pi / \sqrt{t}) d t $$
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Use the information to determine \(\int_{a}^{b} f(x) d x\) and \(\int_{a}^{b} g(x) d x\). $$ \int_{a}^{b}(f(x)+4 g(x)) d x=5, \int_{a}^{b} 3 g(x) d x=-2 $$
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In each of Exercises \(29-34,\) calculate the derivative of \(F(x)\) with respect to \(x\) $$ F(x)=\int_{x}^{\pi} \cos (\sqrt{t}) d t $$
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Calculate the given definite integral. \(\int_{0}^{1} \frac{2^{x}+3^{x}}{5^{x}} d x\)
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