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Magazine Chapter 6: Problem 63
Miscellaneous integrals Evaluate the following integrals. \(\int_{1}^{2 e} \frac{3^{\ln x}}{x} d x\)
Short Answer
Expert verified
Based on the solution provided, find the value of the integral:
\(\int_{1}^{2e} \frac{3^{\ln x}}{x}dx\)
Solution: \(\frac{9\cdot 3^{\ln 2} - 1}{\ln 3}\)
Step by step solution
01
Rewrite the integrand using properties of logarithms
Using the property of logarithms that \(a^{\ln(x)}=x^{\ln(a)}\), the integrand can be rewritten as:
$$
\frac{3^{\ln x}}{x} = \frac{x^{\ln 3}}{x}.
$$
02
Simplify the integrand
Now, divide both terms in the fraction by x, we get:
$$
\frac{x^{\ln 3}}{x} = x^{\ln 3 - 1}.
$$
03
Apply Substitution Method
Let's make a substitution \(u=\ln x\), so \(x=e^{u}\). Also, we need to change the limits of integration accordingly. When \(x=1\), \(u=\ln 1=0\). When \(x=2e\), \(u=\ln(2e)=\ln 2+\ln e = \ln 2+1\). Now, find \(dx\) in terms of \(du\). We have:
$$
\frac{du}{dx} = \frac{1}{x} \Longrightarrow dx = xdu = e^{u}du.
$$
Now, change the integral in terms of \(u\), and substitute \(x=e^u\):
$$
\int_{1}^{2e} x^{\ln 3 - 1}dx = \int_{0}^{\ln 2 + 1} (e^u)^{\ln 3 - 1} e^u du.
$$
04
Simplify the integral and Solve
Simplify the integrand and proceed with integrating the function:
$$
\int_{0}^{\ln 2 + 1} (e^u)^{\ln 3 - 1} e^u du = \int_{0}^{\ln 2 + 1} e^{(\ln 3 - 1)u + u} du = \int_{0}^{\ln 2 + 1} e^{(\ln 3)u} du.
$$
Now integrate the function with respect to \(u\):
$$
\int_{0}^{\ln 2 + 1} e^{(\ln 3)u} du = \frac{1}{\ln 3}e^{(\ln 3)u}\Big|_0^{\ln 2 + 1} = \frac{1}{\ln 3}(e^{(\ln 2 + 1)\ln 3} - e^{0\ln3}).
$$
05
Evaluate the integral
Now, find the value of the integral using the antiderivative found in Step 4:
$$
\frac{1}{\ln 3}(e^{(\ln 2 + 1)\ln 3} - e^{0\ln3}) = \frac{1}{\ln 3}(e^{\ln 2 \ln 3 + \ln 3^2} - e^{0}) = \frac{1}{\ln 3}(2^{\ln 3} \cdot 3^2 - 1).
$$
Since \(2^{\ln 3}=3^{\ln 2}\), we can simplify the answer further:
$$
\frac{1}{\ln 3}(2^{\ln 3} \cdot 3^2 - 1)=\frac{1}{\ln 3}(3^{\ln 2} \cdot 3^2 - 1)=\frac{3^2\cdot 3^{\ln 2} - 1}{\ln{3}}.
$$
So, the integral is equal to \(\int_{1}^{2e} \frac{3^{\ln x}}{x}dx = \frac{3^2\cdot 3^{\ln 2} - 1}{\ln 3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
Integration is a fundamental concept in calculus that involves finding the antiderivative or the area under the curve of a function. It is the opposite operation of differentiation. By integrating a function, you accumulate its value over a specified interval, often to solve problems involving areas, volumes, and other quantities.
In this exercise, we aim to evaluate a definite integral, which calculates the net area under the curve of a function between two values, known as the limits of integration. When evaluating, the focus is on finding the antiderivative first and then applying the limits. The process begins by analyzing the function to be integrated and determining the best method to approach it, such as substitution or partial fractions.
To solve the integral \( \int_{1}^{2e} \frac{3^{\ln x}}{x} dx \), we used the method of rewriting the integrand and applying substitution. This highlights how integration often requires transforming the function into a more manageable form. Understanding the theorem involved in definite integrals and practicing different methods builds a strong foundation in calculus.
In this exercise, we aim to evaluate a definite integral, which calculates the net area under the curve of a function between two values, known as the limits of integration. When evaluating, the focus is on finding the antiderivative first and then applying the limits. The process begins by analyzing the function to be integrated and determining the best method to approach it, such as substitution or partial fractions.
To solve the integral \( \int_{1}^{2e} \frac{3^{\ln x}}{x} dx \), we used the method of rewriting the integrand and applying substitution. This highlights how integration often requires transforming the function into a more manageable form. Understanding the theorem involved in definite integrals and practicing different methods builds a strong foundation in calculus.
Logarithmic Functions
Logarithmic functions involve the logarithm of a number and are the inverse operations of exponential functions. They are crucial in dealing with complex mathematics, often used to simplify expressions and solve equations involving powers and exponents.
In this problem, the integrand \( \frac{3^{\ln x}}{x} \) is initially simplified using properties of logarithms. The property \( a^{\ln(x)}=x^{\ln(a)} \) helps rewrite the function in a more recognizable form: \( \frac{x^{\ln 3}}{x} = x^{\ln 3 - 1} \). This transformation is necessary to make the function integrable. In calculus, the ability to manipulate logarithmic forms is essential as it often surfaces in differential and integral equations.
Moreover, understanding logarithmic properties like the change of base formula, and natural logarithms can allow one to untangle complicated functions that include powers of variables. Logarithmic functions have wide-ranging applications in solving growth and decay problems, making their understanding crucial in both academia and numerous scientific fields.
In this problem, the integrand \( \frac{3^{\ln x}}{x} \) is initially simplified using properties of logarithms. The property \( a^{\ln(x)}=x^{\ln(a)} \) helps rewrite the function in a more recognizable form: \( \frac{x^{\ln 3}}{x} = x^{\ln 3 - 1} \). This transformation is necessary to make the function integrable. In calculus, the ability to manipulate logarithmic forms is essential as it often surfaces in differential and integral equations.
Moreover, understanding logarithmic properties like the change of base formula, and natural logarithms can allow one to untangle complicated functions that include powers of variables. Logarithmic functions have wide-ranging applications in solving growth and decay problems, making their understanding crucial in both academia and numerous scientific fields.
Substitution Method
The substitution method, also known as "u-substitution," is a technique used in integration to simplify and solve integrals more easily. This method involves substituting part of the integrand with a new variable, which turns the original integral into a simpler form.
In our integral, we implemented substitution by setting \( u=\ln x \), transforming \( x=e^u \). This change not only aids in simplifying the algebraic manipulation but also involves adjusting the limits of integration accordingly. When making the substitution, changing the variable often helps manage exponential or logarithmic expressions by reducing them to easier polynomials or exponentials.
To completely apply the substitution method, remember to differentiate your substitution to express \( dx \) in terms of the new variable \( du \). In this exercise, it involved using \( \frac{du}{dx} = \frac{1}{x} \) to find \( dx = e^u du \). This method underscores the interconnectedness of calculus concepts and emphasizes procedural flexibility and mathematical creativity, essential skills in mathematics.
In our integral, we implemented substitution by setting \( u=\ln x \), transforming \( x=e^u \). This change not only aids in simplifying the algebraic manipulation but also involves adjusting the limits of integration accordingly. When making the substitution, changing the variable often helps manage exponential or logarithmic expressions by reducing them to easier polynomials or exponentials.
To completely apply the substitution method, remember to differentiate your substitution to express \( dx \) in terms of the new variable \( du \). In this exercise, it involved using \( \frac{du}{dx} = \frac{1}{x} \) to find \( dx = e^u du \). This method underscores the interconnectedness of calculus concepts and emphasizes procedural flexibility and mathematical creativity, essential skills in mathematics.
