Question

Evaluate the following integrals. $$\int_{1}^{2 e} \frac{3^{\ln x}}{x} d x$$

Short answer

Question: Evaluate the definite integral of the given function:$$\int_{1}^{2 e} \frac{3^{\ln x}}{x} d x$$ Answer: The definite integral of the given function is $$\int_{1}^{2 e} \frac{3^{\ln x}}{x} d x = \frac{(2e)^{\ln 3} - 1}{\ln 3}$$

Step-by-step solution

Recall the properties of exponents and logarithms

Recall that for any positive number \(a\), we have $$a^{\ln x}=x^{\ln a}$$ This property allows us to rewrite the given integrand.

Rewrite the integrand using the property

Using the property from step 1, we can rewrite the integrand as: $$\frac{3^{\ln x}}{x} = \frac{x^{\ln 3}}{x}$$ Now, we simplify the expression: $$\frac{x^{\ln 3}}{x} = x^{\ln 3 - 1}$$

Evaluate the antiderivative

To evaluate the given integral, we need to find the antiderivative of the simplified integrand. Since we have a power function, we can use the power rule: $$\int x^{n} d x = \frac{x^{n+1}}{n+1} + C$$ Applying this rule to our integrand, we have: $$\int x^{\ln 3 - 1} d x = \frac{x^{\ln 3}}{\ln 3} + C$$

Evaluate the definite integral

Now that we have calculated the antiderivative, we can evaluate the definite integral by plugging in the bounds of integration: $$\int_{1}^{2e} x^{\ln 3 - 1} d x = \frac{x^{\ln 3}}{\ln 3} \Big|_{1}^{2 e}$$ Evaluate the antiderivative at the upper and lower bounds: $$\frac{(2e)^{\ln 3}}{\ln 3} - \frac{1^{\ln 3}}{\ln 3}$$

Simplify the result

Simplify the expression to obtain the final result: $$\frac{(2e)^{\ln 3}}{\ln 3} - \frac{1^{\ln 3}}{\ln 3} = \frac{(2e)^{\ln 3} - 1}{\ln 3}$$ So, $$\int_{1}^{2 e} \frac{3^{\ln x}}{x} d x = \frac{(2e)^{\ln 3} - 1}{\ln 3}$$

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, representing the area under a curve between two specified points, called bounds. When you calculate a definite integral, you are essentially finding the signed area between a function's graph and the x-axis from one point to another. In the context of the exercise, the definite integral is given from the bounds 1 to 2e. This means we are interested in the accumulated area of the function \( \frac{3^{\ln x}}{x} \) from \( x = 1 \) to \( x = 2e \).

• The process involves finding the antiderivative (a function whose derivative is the original function) and then evaluating this at the two bounds, subtracting the value at the lower bound from the value at the upper bound.
• The result is often a number or, as in this exercise, an expression that simplifies to a finite value.

Understanding definite integrals is crucial because they can calculate quantities, like area, volume, and other values that express accumulation over an interval.

Antiderivatives

Antiderivatives are the inverse operation of derivatives. If a function \( F(x) \) is an antiderivative of \( f(x) \), then the derivative of \( F(x) \) is \( f(x) \). This relationship is symbolized by \( F'(x) = f(x) \). Antiderivatives are essential in solving integrals, especially definite integrals. In this exercise, we needed the antiderivative of the simplified function \( x^{\ln 3 - 1} \).

• The power rule for antiderivatives states that for any function of the form \( x^n \), the antiderivative is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.
• Applying the power rule enabled us to find \( \int x^{\ln 3 - 1} \, dx = \frac{x^{\ln 3}}{\ln 3} + C \).

Antiderivatives are the pathway to calculating definite integrals. Once you have the antiderivative, evaluate it using the limits of integration to find the total area or accumulated value.

Properties of Logarithms and Exponents

Logarithms and exponents have several properties that simplify expressions and solve integrals. One vital property is that \( a^{\ln b} = b^{\ln a} \). This enables transformations that make integration more feasible. For our exercise, this property allowed the conversion of \( 3^{\ln x} \) into \( x^{\ln 3} \). Here's why these properties are essential:

• They facilitate the rewriting of complex expressions into simpler forms, suitable for integration or differentiation.
• They offer computational shortcuts that can save significant time and effort when evaluating complex integrals.

Using these properties not only simplifies functions but also clarifies their behavior, helping us solve them more efficiently. Understanding how logarithms and exponents interrelate is key to mastering integral calculus.