Question

Compute the definite integrals. Use a graphing utility to confirm your answers. $$ \int_{1}^{e} \ln \left(x^{2}\right) d x $$

Short answer

The value of the definite integral is 2.

Step-by-step solution

Simplify the integrand

The integrand is \( \ln(x^2) \). We know that \( \ln(x^2) = 2\ln(x) \) using the logarithmic identity \( \ln(a^b) = b \ln(a) \). Thus, the integral becomes \( \int_{1}^{e} 2 \ln(x) \, dx \).

Factor out the constant

Since 2 is a constant, we can factor it out of the integral: \( 2 \int_{1}^{e} \ln(x) \, dx \).

Integration by Parts

To solve \( \int \ln(x) \, dx \), we use integration by parts, where \( u = \ln(x) \) and \( dv = dx \). Then \( du = \frac{1}{x} \, dx \) and \( v = x \).

Apply Integration by Parts

Using the formula \( \int u \, dv = uv - \int v \, du \): \[ \int \ln(x) \, dx = x \ln(x) - \int x \left(\frac{1}{x}\right) \, dx = x \ln(x) - \int 1 \, dx = x \ln(x) - x + C \]

Substitute back and evaluate the definite integral

We had \( 2 \int \ln(x) \, dx \), so \( 2[x \ln(x) - x] \). Evaluate from 1 to \( e \): \( 2[e \ln(e) - e - (1 \ln(1) - 1)] = 2[e \cdot 1 - e - (0 - 1)] = 2[e - e + 1] = 2 \).

Confirm with a graphing utility

Use a graphing utility to plot \( y = \ln(x^2) \) and numerically integrate from \( x = 1 \) to \( x = e \). The graphing utility should confirm that the area under the curve is \( 2 \).

Key concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals where the standard methods, like substitution, fall short. The formula for integration by parts is derived from the product rule of differentiation and is given by: \[\int u \, dv = uv - \int v \, du\]Here,

• \( u \) is a function that is easily differentiable, chosen such that its derivative \( du \) simplifies the integral.
• \( dv \) is the remaining part of the integrand that can be directly integrated to obtain \( v \).

In our exercise, applying integration by parts allows us to transform the complex integral of \( \ln(x) \) into simpler components. By letting \( u = \ln(x) \) and \( dv = dx \), we simplify the process.
Using this technique, we derive the expression \( x\ln(x) - x + C \) for \( \int \ln(x) \, dx \). When evaluating definite integrals, this expression is used along with the limits to find the exact area under the curve.

Logarithmic Identities

Logarithmic identities are essential in simplifying complex expressions involving logarithms. One powerful identity is \( \ln(a^b) = b \ln(a) \), which is used to bring exponents in front of the logarithm as a multiplier. In our problem, this identity transforms \( \ln(x^2) \) into \( 2\ln(x) \).This simplification is crucial because it reduces the complexity of the integration challenge we face. By expressing the integrand \( \ln(x^2) \) in a simpler form, we make the integration more straightforward.

• You can multiply and factor constants, which further simplifies integration tasks.
• The simplifying makes it easier to apply techniques such as integration by parts.

Understanding and utilizing these identities takes what could be a difficult integral and makes it manageable, paving the way for effective and accurate calculation.

Graphing Utility

A graphing utility is an indispensable tool in confirming the results of integrals we've solved analytically. These tools include graphing calculators and software like Desmos or GeoGebra. By using a graphing utility, you can visually plot the function and calculate the area under the curve to verify your solution.Our original exercise solution used a graphing utility to validate the definite integral calculation from \( x = 1 \) to \( x = e \) for the function \( y = \ln(x^2) \). When you plot this curve, the tool calculates the area and confirms it equals 2.

• Graphing utilities provide a visual representation, assisting in understanding the behavior of functions and their integrals.
• They offer an alternative numerical method to evaluate areas, which can be particularly helpful when dealing with more complex integrals.

Besides confirming results, these tools are excellent for exploring functions and discovering geometric interpretations of integrals.

Evaluation of Integrals

Evaluating integrals involves calculating the area under the curve that a function defines. For definite integrals, this includes finding the difference between the antiderivatives evaluated at the upper and lower limits. In the context of our problem, after deriving the antiderivative \( 2[x\ln(x) - x] \), we evaluate it from \( x = 1 \) to \( x = e \).This specific evaluation process:

• Substitutes the upper and lower limits into the antiderivative.
• Calculates the difference between these values to determine the integral's value.

In our exercise, performing these steps revealed that the net area under the curve of \( \ln(x^2) \) from 1 to \( e \) is 2. Mastering evaluation techniques is key for accurate computation in both definite and indefinite integrals, ensuring precise results in more complex problems.