Question

Evaluate the following integrals. Include absolute values only when needed. $$\int_{1}^{2 e} \frac{3^{\ln x}}{x} d x$$

Short answer

Question: Evaluate the definite integral $\int_{1}^{2e} \frac{3^{\ln x}}{x} dx$. Answer: The value of the definite integral is $\frac{1}{\ln 3 - 1}[(\ln 3 - 1)(2e - 1)]$.

Step-by-step solution

Simplify the integrand

Using the property of logarithms and exponentials \(a^{\ln b} = b^{\ln a}\), we can rewrite the integrand as follows: $$\frac{3^{\ln x}}{x} = \frac{x^{\ln 3}}{x}$$ Now, divide x from the numerator and denominator: $$\frac{x^{\ln 3}}{x}= x^{\ln 3 - 1}$$

Perform u-substitution

Now, we will perform a u-substitution to find the antiderivative of the integrand. Let \(u = \ln 3 - 1\), then we have: $$\int_{1}^{2 e} x^u d x$$ Differentiating u with respect to x, we get: $$\frac{d u}{d x} = (\ln 3 - 1) x^{u-1}$$ Now we need to find \(dx\) in terms of \(du\): $$d x = \frac{d u}{(\ln 3 - 1) x^{u-1}}$$ Now, we can make the substitution in the integral: $$\int_{1}^{2 e} x^u \cdot \frac{d u}{(\ln 3 - 1) x^{u-1}}$$ Notice that \(x^{u-1}\) will cancel with \(x^u\) in the numerator, leaving us with: $$\frac{1}{\ln 3 - 1} \int_{1}^{2 e} d u$$

Evaluate the definite integral

Now we can integrate with respect to u: $$\frac{1}{\ln 3 - 1} \int_{1}^{2 e} d u = \frac{1}{\ln 3 - 1} [u]_{1}^{2 e}$$ Evaluating the antiderivative at the limits of integration, we get: $$\frac{1}{\ln 3 - 1}[(\ln 3 - 1)(2e - 1)]$$

Write the final answer

So, the value of the definite integral is: $$\int_{1}^{2 e} \frac{3^{\ln x}}{x} dx = \frac{1}{\ln 3 - 1}[(\ln 3 - 1)(2e - 1)]$$

Key concepts

U-Substitution

U-substitution is a technique used in integration similar to the reverse process of the chain rule in differentiation. It is particularly useful when dealing with composite functions, where the integrand is a function of another function. Imagine you are trying to navigate through a dense forest.
Just as you would look for a clearer path to make your journey easier, u-substitution provides a simpler way to approach complex integrals by changing variables.

For instance, if you have an integral of the form \( \int f(g(x))g'(x) \,dx \), you can let \( u = g(x) \) and thus \( du = g'(x)dx \), effectively transforming the integral into \( \int f(u) \, du \), which is often easier to evaluate. Remember to adjust the limits of integration if you are dealing with a definite integral. In the exercise, the substitution \( u = \ln 3 - 1 \) is chosen because it simplifies the power of \( x \) in the integrand, making the integral more straightforward to evaluate.

Antiderivative

An antiderivative is a function that, when differentiated, yields the original function. Think of it as retracing your steps back home after a walk in the park.
When you integrate a function, you are finding all possible antiderivatives of that function.

In the context of definite integrals, finding the antiderivative allows you to evaluate the integral over a specific interval. You can consider the process of integration as accumulating small pieces to find the whole, similar to piecing together a puzzle. The key step in solving the exercise is recognizing that \( \int x^u dx \) can be addressed by finding its antiderivative, which in a power function situation is \( \frac{x^{u+1}}{u+1} \), given that \( u eq -1 \) to avoid division by zero. Remember, the constant of integration is omitted in definite integrals since it cancels out when evaluating the limits.

Logarithmic Integration

Logarithmic integration involves integrating functions that result in a logarithmic function as part of the antiderivative.
It's akin to unearthing an ancient relic while digging; the artifact being the log function hidden within the integral.

In our problem, logarithmic integration is not explicitly required, but understanding the logarithmic function is crucial to simplifying the integrand. The original integrand \( \frac{3^{\ln x}}{x} \) can be re-expressed using properties of logs, transforming it into a form that is amenable to standard integration techniques. The relationship \( a^{\ln b} = b^{\ln a} \) is what makes this simplification possible. Logarithmic integration is generally used when the integrand is in the form of \( \frac{1}{x} \) or directly involves logarithmic expressions.

Integral Evaluation

Integral evaluation is the process of finding the numerical value of an integral, often referred to as solving the integral. Consider it as checking the answer at the back of a mystery novel.
An integral can represent an area, a volume, or other physical concepts, depending on the context.

In definite integrals, evaluating means calculating the difference in the antiderivative's values at the upper and lower limits of integration. In our exercise, after simplification and employing u-substitution, the integral is reduced to \( \frac{1}{\ln 3 - 1} \int_{1}^{2e} du \), which is straightforward to evaluate. The final step involves substituting the limits into the antiderivative and subtracting to get the result. This process transforms the problem from finding the 'sum' of an infinite series of products into a simple arithmetic operation.