Problem 36 - Calculus Single Variable

Question

Use the method of substitution to evaluate the definite integrals. $$ \int_{1}^{e} \frac{\sqrt{\ln (x)}}{x} d x $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{2}{3} \).

Step 1: Choose the substitution variable

We start by performing the substitution. Let us consider the substitution \( u = \ln(x) \). Notice that with this substitution, the differential changes as well. We have \( du = \frac{1}{x} dx \).

Step 2: Change the limits of integration

Now we substitute the limits of the integral. When \( x = 1 \), then \( \ln(1) = 0 \), so the lower limit is 0. When \( x = e \), \( \ln(e) = 1 \), so the upper limit is 1. So the integral becomes \( \int_{0}^{1} \sqrt{u} \, du \).

Step 3: Evaluate the new integral

The integral now is \( \int_{0}^{1} u^{1/2} \, du \). The antiderivative of \( u^{1/2} \) is \( \frac{2}{3} u^{3/2} \).

Step 4: Apply the limits of integration

Substitute the limits 0 and 1 into the antiderivative. First, evaluate at the upper limit: \( \frac{2}{3} (1)^{3/2} = \frac{2}{3} \). Then, evaluate at the lower limit: \( \frac{2}{3} (0)^{3/2} = 0 \). So, the integral evaluates to \( \frac{2}{3} - 0 \).

Step 5: Simplify the result

The definite integral evaluates to \( \frac{2}{3} \).

Key Concepts

Definite Integrals

Definite integrals help us with finding the area under a curve over a specific interval. It is more than computing an area, it involves understanding how the curve behaves across the interval. In simple terms, a definite integral lets us find the accumulated value of a function from one point to another. In our example, the given definite integral is \( \int_{1}^{e} \frac{\sqrt{\ln (x)}}{x} \ dx \) which means we look at the function within the limits 1 and \( e \).
The notation \( \int_{a}^{b} f(x) \, dx \) describes a process where we calculate this accumulation from point \( a \) to point \( b \).
The key significance of definite integrals is that they provide both limits, thus allowing for precise calculation of areas or accumulated quantities between these points.

Substitution Method

The substitution method is like changing the view of our problem to make it simpler to solve. Sometimes the given integral is too complex directly. By substitution, we alter variables to get an easier form.
In our example, we used the substitution \( u = \ln(x) \). This changed the form of the integral to \( \int_{0}^{1} \sqrt{u} \, du \), a form that's easier to handle.
Working with this method usually involves:

• Choosing a substitution that simplifies the integral.
• Changing the differential accordingly, here \( du = \frac{1}{x} \, dx \).
• Adjusting the limits of integration to reflect the new variable, from \( (1, e) \) to \( (0, 1) \).

The substitution essentially focuses on simplifying the integrand into something more familiar or easier to evaluate.

Limits of Integration

When dealing with definite integrals, the limits of integration indicate where we start and end the calculation. They define the boundaries of the area or accumulation we measure. For the given integral \( \int_{1}^{e} \frac{\sqrt{\ln (x)}}{x} \, dx \), the original limits \( x = 1 \) to \( x = e \) meant we evaluated the function from 1 to \( e \).
After the substitution, the limits changed too. These became \( u = 0 \) for \( x = 1 \), and \( u = 1 \) for \( x = e \).
The alteration of limits helps us keep the integrity of the integral even if we change the variable. Always ensure to adjust limits when employing substitution to ensure the results reflect the correct interval.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse process of differentiation. To calculate an integral, we essentially find the antiderivative of the function. In our example, this was transforming \( u^{1/2} \) into its antiderivative \( \frac{2}{3}u^{3/2} \).
Once we have the antiderivative, we evaluate it with the limits of integration.
The steps generally involve:

• Finding the antiderivative of the simplified integrand.
• Applying the limits to evaluate the definite integral.
• Subtracting the lower limit value from the upper limit value as done with \( \frac{2}{3} - 0 \), which results in \( \frac{2}{3} \).

Antiderivatives form a crucial part of calculating definite integrals and understanding how the area or accumulation changes over a specified interval.