Problem 102 - Calculus and its applications

Question

Explain the error that has been made in each. $$ \begin{aligned} \int_{1}^{2}\left(\ln x-e^{x}\right) d x &=\left[\frac{1}{x}-e^{x}\right]_{1}^{2} \\ &=\left(\frac{1}{2}-e^{2}\right)-\left(1-e^{1}\right) \\ &=e-e^{2}-\frac{1}{2} \end{aligned} $$

Step-by-Step Solution

Answer

The error was using the wrong antiderivative for \( \ln x \), which should be \( x \ln x - x \).

Step 1: Analyze the Original Integral

The given integral is \( \int_{1}^{2}(\ln x - e^{x}) \, dx \). The goal is to find the antiderivative of each component \( \ln x \) and \(-e^x \), and evaluate it from 1 to 2.

Step 2: Antidifferentiation

The antiderivative of \( e^x \) is \( e^x \), so \(-e^x\) becomes \(-e^x\). However, the correct antiderivative for \( \ln x \) is \( x \ln x - x \), not \( \frac{1}{x} \). This is the first error in the solution.

Step 3: Correct Antiderivative Expression

The correct antiderivative expression should be \([x \ln x - x - e^x]_1^2 \). This will represent the integral of both terms from 1 to 2.

Step 4: Evaluate the Correct Antiderivative

Plug in the upper limit 2 into \(x \ln x - x - e^x\) to get \(2 \ln 2 - 2 - e^2\). Then plug in the lower limit 1 to get \(1 \ln 1 - 1 - e^1 = -1 - e\).

Step 5: Simplify the Evaluation

Subtract the lower limit results from the upper limit: \((2 \ln 2 - 2 - e^2) - (-1 - e) = 2 \ln 2 - 1 - e^2 + e\).

Key Concepts

Antiderivatives

Antiderivatives, also known as indefinite integrals, are essentially the reverse process of differentiation. When you find an antiderivative of a function, it means you are looking for a new function whose derivative is the original function. This can be visualized as the opposite of taking the derivative.

In the context of this integral, we are trying to find the antiderivative of both components \( \ln x \) and \(-e^x \). However, the original attempt made an error by assuming \( \ln x \) has the antiderivative \( \frac{1}{x} \).

• The antiderivative of \( \ln x \) is \( x \ln x - x \).
• The antiderivative of \( -e^x \) is still \( -e^x \).

Understanding the correct antiderivative is crucial for solving integrals properly. This is because the integral itself represents an accumulation of values described by the function, and the antiderivative helps capture this concept in a broader mathematical form. By rewriting the integral in the antiderivative context, you can reveal all these accumulated differences across a given range.

Definite Integrals

Definite integrals provide a precise numerical value that represents the net area under a curve within a given interval. Unlike indefinite integrals that include an arbitrary constant \( C \), definite integrals have specific limits where the function is evaluated.

To compute a definite integral:

• Determine the antiderivative or primitive function of the integrand.
• Evaluate this antiderivative at the upper and lower limits of integration.
• Subtract the evaluated lower limit result from the upper limit result.

For the exercise in question, after correctly finding the antiderivative \( [x \ln x - x - e^x]_1^2 \), calculate it at \( x = 2 \) and \( x = 1 \). The difference between these values will give the actual value of the integral from 1 to 2.

Keep in mind that definite integrals give significant insights into real-world problems, such as finding total displacement or total area, by summing infinitely small quantities over a range.

Logarithmic Integration

Logarithmic integration comes into play when the integrand includes functions like \( \ln x \). It is a special case in integral calculus because the derivative of \( \ln x \) is quite unlike typical polynomial functions.

When computing the integral of \( \ln x \), it's useful to recognize it requires integration by parts, a technique especially handy for products of functions. For the function \( \ln x \), we choose:

• \( u = \ln x \)
• \( dv = dx \)

Then, applying integration by parts:\[\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C \]
In our original problem, failure to correctly find the antiderivative of \( \ln x \) led to an erroneous evaluation of the integral. Always ensure that logarithmic functions are handled with the proper methods like integration by parts to avoid mistakes.