Problem 96 - Calculus Early Transcendentals

Question

Compute \(\int_{0}^{1} \ln x d x\) using integration by parts. Then explain why \(-\int_{0}^{\infty} e^{-x} d x\) (an easier integral) gives the same result.

Step-by-Step Solution

Answer

Answer: Both integrals give the same result because when computed using appropriate methods, they both evaluate to 1. The first integral is calculated using integration by parts, while the second one is a standard integral which can be computed more easily. Despite the difference in techniques employed, the final results match, demonstrating that the easier integral provides the same result as the more complicated one.

Step 1: Apply Integration by Parts Formula

Integration by parts follows the formula: $$\int u\ dv = uv - \int v\ du$$ where u and dv are functions of x. For our integral \(\int_{0}^{1} \ln x dx\), we'll choose the following functions: u = \(\ln x\) and dv = \(dx\) Now we need to find du and v; so let's differentiate u and integrate dv.

Step 2: Differentiate u and Integrate dv

To find du, differentiate u with respect to x: du = \(\frac{d(\ln x)}{dx} = \frac{1}{x} dx\) To find v, integrate dv with respect to x: v = \(\int dx = x\)

Step 3: Plug u, v, du, and dv into the Integration by Parts Formula

Now, we have: u = \(\ln x\) dv = \(dx\) du = \(\frac{1}{x} dx\) v = \(x\) Plugging these into the integration by parts formula, we get: $$\int_{0}^{1} \ln x\ dx = \int_{0}^{1} u\ dv = uv|_{0}^{1} - \int_{0}^{1} v\ du = x \ln x |_{0}^{1} - \int_{0}^{1} x\left(\frac{1}{x}\right) dx$$ Simplify the integral: $$x \ln x |_{0}^{1} - \int_{0}^{1} dx$$

Step 4: Evaluate the Resulting Integrals

Evaluate the first part of the equation: $$x \ln x |_{0}^{1} = (1 \cdot \ln 1) - (0 \cdot \ln 0) = 0$$ Note that the second term \(0 \cdot \ln 0\) would be undefined, but here we consider it as a limit: $$\lim_{x \to 0} x \ln x = 0$$ Now, evaluate the remaining integral: $$-\int_{0}^{1} dx = -[x]_0^1 = -(1 - 0) = -1$$ So, the value of \(\int_0^1 \ln x\ dx = 0 - (-1) = 1\) Now, let's solve the easier integral:

Step 5: Solve the Easier Integral

The easier integral is given by: $$-\int_{0}^{\infty} e^{-x} dx$$ This is a standard integral and can be computed easily: $$-\left[-e^{-x} \right]_{0}^{\infty} = -(-1 + 1) = 1$$

Step 6: Compare Results

Both integrals have the same value: \(\int_{0}^{1} \ln x dx = 1\) \(-\int_{0}^{\infty} e^{-x} dx = 1\) Thus, we have demonstrated that the easier integral gives the same result as the original, more complicated integral, as required by the exercise.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, utilized to determine the area under a curve over a specific interval. In the context of this exercise, we are focusing on the interval from 0 to 1. This is represented by the notation \(\int_{0}^{1}\).

Understanding definite integrals involves two main steps:

• Firstly, we compute the antiderivative, or the "indefinite integral," of the function. This is a formula that can produce all possible antiderivatives.
  
• Secondly, applying the limits (in this case, 0 and 1), we evaluate the calculated antiderivative at both ends of the interval.
  

The result of substituting the upper limit and subtracting the value obtained by substituting the lower limit provides the exact area" under the curve between these bounds. In the solution, this process is demonstrated by using integration by parts.

Limit Evaluation

When dealing with definite integrals, especially those that involve challenging boundaries (like 0 and positive or negative infinity), it's crucial to incorporate limit evaluations.

In the solution we reviewed, you notice a term \(x \ln x\) where \(x = 0\), which suggests an undefined form. To handle such cases, we use limits to carefully analyze the behavior of the function as the variable approaches a particular point. This is seen through the expression \(\lim_{x \to 0} x \ln x = 0\).

Here's a brief reminder on handling limits in integrals:

• Identify the undefined behavior, such as \(x\ln x\) as \(x\) approaches zero.
  
• Apply limit rules and l'Hôpital's Rule if necessary to determine the behavior as the variable approaches the troublesome value.
  
• Establish that the expression simplifies to a number, allowing the integral evaluation to proceed smoothly.
  

This approach is essential for properly evaluating integrals with limits that result in complex or indeterminate forms.

Natural Logarithm Integration

Integrating natural logarithms like \(\ln x\) takes a special approach, often requiring integration by parts. This strategy simplifies the process by breaking the integral into more manageable pieces.

Here's a recap:

• Identify parts: Select \(u\) as \(\ln x\) (since it simplifies when differentiated), and let \(dv = dx\), making the integral \(\int u\ dv\) straightforward.
  
• Compute \(du\) and \(v\): Differentiate \(u\) to get \(du = \frac{1}{x} dx\), and integrate \(dv\) to obtain \(v = x\).
  
• Substitute into the formula: Use the integration by parts formula \(\int u\ dv = uv - \int v\ du\), resulting in a simplified expression to evaluate further.
  

The integral of \(\ln x\) can be tricky because it doesn’t directly lead to a neat elementary function through standard antiderivatives. However, by utilizing integration by parts, we break the task into elementary integrals that are much simpler to evaluate.