Problem 24
Question
Compute the definite integral and interpret the result in terms of areas. $$ \int_{1}^{4}(x-3 \ln x) d x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \(16.5 - 12 \ln 4\), representing the signed area of the curve \(y = x - 3 \ln x\) between x=1 and x=4.
1Step 1: Understand the Integral
The given problem asks for the computation of the definite integral \( \int_{1}^{4}(x-3 \ln x) \, dx \). This requires us to find the function whose derivative is the integrand \( x - 3 \ln x \), evaluate this antiderivative at the upper and lower limits, and find the difference between these values.
2Step 2: Antiderivative Calculation
Determine the antiderivative of the integrand \( x - 3 \ln x \). The antiderivative of \( x \) is \( \frac{x^2}{2} \), and for \( -3 \ln x \), use the substitution method or refer to tables to conclude it is \(-3x \ln x + 3x\) (derived using integration by parts). Thus, the antiderivative overall is \( \frac{x^2}{2} - 3x \ln x + 3x \).
3Step 3: Apply the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, evaluate the antiderivative obtained in Step 2 at the upper limit (4) and lower limit (1), and subtract: \[ F(4) - F(1) = \left( \frac{4^2}{2} - 3 \times 4 \ln 4 + 3 \times 4 \right) - \left( \frac{1^2}{2} - 3 \times 1 \ln 1 + 3 \times 1 \right) \] Simplify using the property \( \ln 1 = 0 \).
4Step 4: Simplify the Evaluation
Compute the values step by step:1. Evaluate the expression at 4: \( \frac{16}{2} - 12 \ln 4 + 12 \) = 8 - 12 \ln 4 + 12 = 20 - 12 \ln 4.2. Evaluate the expression at 1: \( \frac{1}{2} + 3 \), since \( \ln 1 = 0 \), which equals 3.5.3. Subtract these results: 20 - 12 \ln 4 - 3.5 = 16.5 - 12 \ln 4.
5Step 5: Interpret the Result
The result \( 16.5 - 12 \ln 4 \) represents the signed area between the curve \( y = x - 3 \ln x \) and the x-axis from x=1 to x=4. If \( 12 \ln 4 \) is less than 16.5, this area is above the x-axis; otherwise, it includes areas below the axis as negatives.
Key Concepts
Fundamental Theorem of CalculusAntiderivative CalculationIntegration by Parts
Fundamental Theorem of Calculus
The fundamental theorem of calculus is a key principle that connects the concepts of differentiation and integration. It comprises two main parts:
The first part provides a way to compute an integral by finding an antiderivative. It states that if you have a continuous function, its definite integral over an interval can be found using its antiderivative.
In simple terms, if you know the slope (or derivative) of a line, and you want to know the area under its curve over a specific interval, you can use the antiderivative of that function to find it.
The second part shows that the derivative and the integral are inverse processes. It suggests that if you take the derivative of an integral function, you end up with the original function that was integrated.
This theorem makes it possible to evaluate definite integrals quickly once you find the antiderivative.
In our problem, after finding the antiderivative of the function \(x - 3 \ln x\), the fundamental theorem of calculus allows us to use the values of this antiderivative at the boundaries 1 and 4 to find the solution to the definite integral.
The first part provides a way to compute an integral by finding an antiderivative. It states that if you have a continuous function, its definite integral over an interval can be found using its antiderivative.
In simple terms, if you know the slope (or derivative) of a line, and you want to know the area under its curve over a specific interval, you can use the antiderivative of that function to find it.
The second part shows that the derivative and the integral are inverse processes. It suggests that if you take the derivative of an integral function, you end up with the original function that was integrated.
This theorem makes it possible to evaluate definite integrals quickly once you find the antiderivative.
In our problem, after finding the antiderivative of the function \(x - 3 \ln x\), the fundamental theorem of calculus allows us to use the values of this antiderivative at the boundaries 1 and 4 to find the solution to the definite integral.
Antiderivative Calculation
Finding the antiderivative is like working backward from a derivative to find the original function before differentiation took place. It is sometimes called 'integration'.
For the function \(x - 3 \ln x\), we calculate the antiderivative by dealing with each term separately:
This gives us a complete expression to work with when applying the fundamental theorem of calculus.
For the function \(x - 3 \ln x\), we calculate the antiderivative by dealing with each term separately:
- For \(x\), the antiderivative is \(\frac{x^2}{2}\). This follows from the power rule for integration.
- The antiderivative for \(-3 \ln x\) is more complex. It involves using a technique called integration by parts.
This gives us a complete expression to work with when applying the fundamental theorem of calculus.
Integration by Parts
Integration by parts is a technique derived from the product rule of differentiation. It is particularly useful when dealing with products of functions.
This method requires choosing which parts of the function to differentiate and integrate. In our case of \(-3 \ln x\), set
By understanding integration by parts, we can tackle integrals involving products of functions more effectively. This technique enriches our toolkit when calculating integrals, especially those akin to \(-3 \ln x\).
This method requires choosing which parts of the function to differentiate and integrate. In our case of \(-3 \ln x\), set
- \(u = \ln x\), so \(du = \frac{1}{x}dx\)
- \(dv = -3dx\), so \(v = -3x\)
By understanding integration by parts, we can tackle integrals involving products of functions more effectively. This technique enriches our toolkit when calculating integrals, especially those akin to \(-3 \ln x\).
Other exercises in this chapter
Problem 23
Compute the definite integral and interpret the result in terms of areas. $$ \int_{1}^{4} \frac{x^{2}-3}{x} d x $$
View solution Problem 24
Use a calculator or computer to evaluate the integral. $$ \int_{1.1}^{1.7} 10(0.85)^{t} d t $$
View solution Problem 25
Use a calculator or computer to evaluate the integral. $$ \int_{1}^{2} 2^{x} d x $$
View solution Problem 25
Compute the definite integral \(\int_{0}^{4} \cos \sqrt{x} d x\) and interpret the result in terms of areas.
View solution