Problem 69
Question
Evaluate the definite integral. \( \displaystyle \int^{e^4}_{e} \frac{dx}{x \sqrt{\ln x}} \)
Step-by-Step Solution
Verified Answer
The definite integral evaluates to 2.
1Step 1: Substitution
To solve the integral \( \int \frac{dx}{x \sqrt{\ln x}} \), we first use the substitution method. Let \( u = \ln x \), which implies that \( du = \frac{1}{x}dx \). Therefore, the integral becomes \( \int \frac{du}{\sqrt{u}} \).
2Step 2: Change the Limits of Integration
We need to adjust our limits of integration for the substitution. When \( x = e \), the lower limit, \( u = \ln e = 1 \). When \( x = e^4 \), the upper limit, \( u = \ln e^4 = 4 \). Thus, the integral changes to \( \int_{1}^{4} u^{-1/2} \, du \).
3Step 3: Integrate
Integrate \( \int u^{-1/2} \, du \). The antiderivative of \( u^{-1/2} \) is \( 2u^{1/2} \). So, the integral becomes \( 2u^{1/2} \).
4Step 4: Evaluate the Definite Integral
Evaluate \( 2u^{1/2} \) from 1 to 4. This gives \( 2(4^{1/2}) - 2(1^{1/2}) \). Calculate these values to find \( 2 \times 2 - 2 \times 1 = 4 - 2 = 2 \).
Key Concepts
Substitution MethodLimits of IntegrationAntiderivativeEvaluation of Definite Integrals
Substitution Method
The substitution method is a technique often used to simplify the process of finding integrals, especially when dealing with complicated expressions. In the context of definite integrals, substitution is useful for transforming the integral into a more manageable form.
In this exercise, the substitution involved letting \( u = \ln x \). This choice is strategic since it turns the more complex expression \( \frac{1}{x\sqrt{\ln x}} \) into a simpler one. By differentiating \( u = \ln x \), we find that \( du = \frac{1}{x} dx \). This relation allows us to rewrite our original integral in terms of \( u \) as \( \int \frac{du}{\sqrt{u}} \).
The substitution method simplifies the integration process, enabling us to work with a simpler integrand. This results in a new integral to solve, with the substitution essentially acting as a tool to bridge the gap between a complex integral and a straightforward one.
In this exercise, the substitution involved letting \( u = \ln x \). This choice is strategic since it turns the more complex expression \( \frac{1}{x\sqrt{\ln x}} \) into a simpler one. By differentiating \( u = \ln x \), we find that \( du = \frac{1}{x} dx \). This relation allows us to rewrite our original integral in terms of \( u \) as \( \int \frac{du}{\sqrt{u}} \).
The substitution method simplifies the integration process, enabling us to work with a simpler integrand. This results in a new integral to solve, with the substitution essentially acting as a tool to bridge the gap between a complex integral and a straightforward one.
Limits of Integration
After using substitution, it's crucial to adjust the limits of integration from the original variable to the new variable. This ensures that the definite integral remains correct in its new form. When we set \( u = \ln x \), we also change our limits of integration based on this substitution.
Originally, our limits were from \( x = e \) to \( x = e^4 \). By substituting these values into \( u = \ln x \), we convert the limits. For \( x = e \), the lower limit becomes \( u = \ln e = 1 \). Similarly, for \( x = e^4 \), the upper limit becomes \( u = \ln e^4 = 4 \). Thus, the new limits of integration are from 1 to 4.
Updating the limits correctly is an important step. It ensures that the area being calculated is the same before and after substitution, maintaining the integrity of the definite integral.
Originally, our limits were from \( x = e \) to \( x = e^4 \). By substituting these values into \( u = \ln x \), we convert the limits. For \( x = e \), the lower limit becomes \( u = \ln e = 1 \). Similarly, for \( x = e^4 \), the upper limit becomes \( u = \ln e^4 = 4 \). Thus, the new limits of integration are from 1 to 4.
Updating the limits correctly is an important step. It ensures that the area being calculated is the same before and after substitution, maintaining the integrity of the definite integral.
Antiderivative
An antiderivative represents a function whose derivative is the integrand. Finding the antiderivative is a key part of the integration process.
In our example, after substitution, we needed to find the antiderivative of \( u^{-1/2} \). The rule for integrating powers of \( u \) states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) for \( n eq -1 \). Applying this rule, the antiderivative of \( u^{-1/2} \) is \( 2u^{1/2} \).
The antiderivative now represents a function that we can evaluate from the new limits of 1 to 4. This result reflects the accumulation of the area under the curve for the transformed integrand, which is crucial for evaluating the definite integral.
In our example, after substitution, we needed to find the antiderivative of \( u^{-1/2} \). The rule for integrating powers of \( u \) states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \) for \( n eq -1 \). Applying this rule, the antiderivative of \( u^{-1/2} \) is \( 2u^{1/2} \).
The antiderivative now represents a function that we can evaluate from the new limits of 1 to 4. This result reflects the accumulation of the area under the curve for the transformed integrand, which is crucial for evaluating the definite integral.
Evaluation of Definite Integrals
Evaluating a definite integral involves computing the difference of the antiderivative evaluated at the upper limit and the lower limit. This evaluation gives the net area between the curve and the horizontal axis over the given interval.
In our scenario, once we found the antiderivative \( 2u^{1/2} \), we executed this process from the limits of 1 to 4. We calculated \( 2(4^{1/2}) - 2(1^{1/2}) \), which simplifies to \( 2 \times 2 - 2 \times 1 \), resulting in 2.
Thus, the evaluation of the definite integral confirms the total area under the curve from the transformed limits, providing the final solution to the original integration problem.
In our scenario, once we found the antiderivative \( 2u^{1/2} \), we executed this process from the limits of 1 to 4. We calculated \( 2(4^{1/2}) - 2(1^{1/2}) \), which simplifies to \( 2 \times 2 - 2 \times 1 \), resulting in 2.
Thus, the evaluation of the definite integral confirms the total area under the curve from the transformed limits, providing the final solution to the original integration problem.
Other exercises in this chapter
Problem 68
If \( \displaystyle f(x) = \int^{\sin x}_0 \sqrt{1 + t^2} \, dt \) and \( \displaystyle g(y) = \int^y_3 f(x) \, dx \), find \( g''(\pi/6) \).
View solution Problem 68
Which of the integrals \( \displaystyle \int^{0.5}_0 \cos (x^2) \,dx \), \( \displaystyle \int^{0.5}_0 \cos \sqrt{x} \,dx \) is larger? Why?
View solution Problem 69
If \( f(1) = 12 \), \( f' \) is continuous, and \( \displaystyle \int^4_1 f'(x) \, dx = 17 \), what is the value of \( f(4) \)?
View solution Problem 70
Evaluate the definite integral. \( \displaystyle \int^2_0 (x - 1)e^{(x - 1)^2} \, dx \)
View solution