Problem 32 - Calculus Single Variable

Question

Determine whether the given improper integral converges or diverges. If it converges, then evaluate it. $$ \int_{-\infty}^{0} e^{-x} d x $$

Step-by-Step Solution

Answer

The integral converges and evaluates to -1.

Step 1: Write the Integral in Limit Form

To determine the convergence of an improper integral with \[-\infty\] as its lower limit, express the integral using limits. Replace \(-\infty\) with a variable, say \(a\), and take the limit as \(a\) approaches \(-\infty\). \[ \int_{-\infty}^{0} e^{-x} dx = \lim_{a \to -\infty} \int_{a}^{0} e^{-x} dx \]

Step 2: Find the Indefinite Integral

Compute the indefinite integral of \(e^{-x}\). The antiderivative of \(e^{-x}\) is \(-e^{-x}\): \[ \int e^{-x} dx = -e^{-x} + C \] We can now use this to evaluate the definite integral.

Step 3: Evaluate the Definite Integral

Using the antiderivative from Step 2, evaluate \(\int_{a}^{0} e^{-x} dx\) and apply the limits: \[ \int_{a}^{0} e^{-x} dx = \left[-e^{-x}\right]_{a}^{0} = \left(-e^{0}\right) - \left(-e^{a}\right) = -1 + e^{a} \]

Step 4: Apply the Limit

Now, apply the limit as \(a\) approaches \(-\infty\):\[ \lim_{a \to -\infty} (-1 + e^{a}) \] Since \( e^{a} \to 0 \) as \( a \to -\infty \), the expression simplifies to:\[ -1 + 0 = -1 \]

Step 5: Conclusion on Convergence

Because the limit exists and results in a real number, the improper integral converges.The integral evaluates to \(-1\).

Key Concepts

Convergence and Divergence of Integrals

When evaluating improper integrals, one of the primary tasks is to determine whether they converge or diverge. An improper integral can be thought of as converging if it results in a finite number. Otherwise, it diverges, meaning it approaches infinity or does not settle on any specific value. In this case, the exercise uses the improper integral with a lower limit of negative infinity:

• Express the integral using limits to assess convergence.
• Replace the infinity limits with a variable, for example, a, and calculate the limit of the integral as a approaches infinity or negative infinity.

In this example, the integral converges means it results in a finite number, due to each part of the solution cleanly resulting in a real number after calculating the limit.

Indefinite Integrals

An indefinite integral helps us find the antiderivative of a function and is crucial in solving definite integrals later. In this exercise, you need to find the indefinite integral before evaluating the improper integral specifically. For the function

• The antiderivative of \( e^{-x} \) is necessary. It is solved as \(-e^{-x} + C\), where \( C \) is an arbitrary constant.

The indefinite integration step is crucial because it simplifies the process of finding the solution to the given definite or improper integral. This step essentially prepares the function for limit evaluation.

Limit Evaluation

Limit evaluation plays a central role in determining the convergence of an improper integral. After defining the integral in terms of limits, you'll apply limits to solidify the solution. Here is how it is done:

• Substitute the limits back into the antiderivative calculated in the indefinite integral step.
• For \( \lim_{a \to -\infty} (-1 + e^a) \), the expression \( e^a \to 0 \) as \( a \to -\infty \).

Evaluating this limit provides a concrete answer to the convergence question, as the expression simplifies to \(-1 + 0 = -1\), ensuring a finite value indicates convergence.

Definite Integrals

Definite integrals involve evaluating the antiderivative at the upper and lower bounds of the function. For improper integrals especially, this incorporates the limits:

• First, determine the indefinite integral to establish the antiderivative.
• Next, evaluate the bounds—substitute the limits and compute their outcomes.

In the exercise, this is represented by solving \( \int_{a}^{0} e^{-x} dx \), then inserting \( a \) and \( 0 \) into \(-e^{-x} + C\), leading to \([-e^0] - [-e^a]\) or \(-1 + e^a\). Proper evaluation of these steps helps conclude the integral effectively for an end value linked directly to limit evaluation.