Problem 5 - Calculus and its applications

Question

Determine whether each improper integral is convergent or divergent, and find its value if it is convergent. $$ \int_{0}^{\infty} 3 e^{-3 x} d x $$

Step-by-Step Solution

Answer

The integral is convergent, and its value is 1.

Step 1: Identify the type of improper integral

The given integral \( \int_{0}^{\infty} 3 e^{-3x} \, dx \) is an improper integral because the upper limit of integration is infinity. We need to test for convergence or divergence.

Step 2: Set up the limit definition

To determine the convergence, express the integral as a limit: \[ \lim_{b \to \infty} \int_{0}^{b} 3e^{-3x} \, dx \]. This allows us to evaluate the integral from 0 to \( b \), and then examine the behavior as \( b \) approaches infinity.

Step 3: Evaluate the integral from 0 to b

Calculate \( \int 3e^{-3x} \, dx \). Use the substitution method: let \( u = -3x \), hence \( du = -3 \, dx \), so \( dx = -\frac{1}{3} \, du \). Thus, \( \int 3e^{-3x} \, dx = \int -e^{u} \, du = -e^{-3x} + C \).

Step 4: Apply the limits of integration

Substitute back into the original integral's limit setup: \[ -e^{-3x} \bigg|_{0}^{b} = (-e^{-3b}) - (-e^{0}) = -e^{-3b} + 1 \].

Step 5: Evaluate the limit as b approaches infinity

Calculate \[ \lim_{b \to \infty} (-e^{-3b} + 1) \]. As \( b \to \infty \), \( e^{-3b} \to 0 \). Thus, \( -e^{-3b} + 1 \to 1 \).

Step 6: Determine convergence status

Since the limit exists and is equal to 1, the improper integral \( \int_{0}^{\infty} 3e^{-3x} \, dx \) is convergent. The value of the integral is 1.

Key Concepts

Convergence

When dealing with improper integrals, one of the key concepts is determining whether the integral converges or diverges. Convergence means that as we evaluate the integral, the result approaches a specific value. In mathematical terms, an integral is convergent if the limit of the integral's expression, as one of its bounds approaches infinity, exists and is finite. Consider the problem \[ \int_{0}^{\infty} 3 e^{-3x} \, dx \] This integral converges because we can evaluate the limit of the integral and obtain a finite number, which is 1, as noted in the solution. Key reasons why convergence happens include:

• The integral's function decreases rapidly enough as the variable increases.
• In this case, the factor \(e^{-3x}\) decreases exponentially as \(x\) increases, causing the area under the curve from the lower bound to infinity to become finite.

To test for convergence: 1. Set up the integral as a limit problem.2. Find the antiderivative, apply the limits, and calculate the limit.3. If the limit results in a definite number, the integral is said to converge.

Divergence

Divergence is the opposite of convergence and occurs when the integral does not settle on a specific value as it is evaluated. In simpler terms, for a divergent integral, as we approach infinity (or some other improper bound), there is no finite limit; the value grows without bound or oscillates indefinitely. Understanding whether an improper integral diverges can involve:

• Estimating the behavior of the function being integrated as it approaches infinity or its boundary points.
• Considering whether the rate of increase or decrease in the function is sufficiently slow or fast, respectively.

Let's say our integral did not have the \(e^{-3x}\) but another function, like \(3x\). In such cases, \[ \int_{0}^{\infty} 3x \, dx \] would diverge because its integral leads to an infinite limit as the boundary approaches infinity. Remember, you need to establish a pattern as to whether it approaches a specific number or not when deciding on convergence or divergence for an integral.

Integration by Substitution

Integration by substitution is a powerful technique, especially handy with integrals that involve composite functions or where direct integration isn't straightforward. Here's how substitution can simplify solving integrals:

• Replaces a complex part of the integrand with a new variable, simplifying the integral.
• By selecting an appropriate substitution, one can transform the integral into a simpler form that is easier to evaluate.

In the example \(\int 3e^{-3x} \, dx\), using substitution can drastically simplify the integration:1. Pick a substitution: let \(u = -3x\).2. Differentiate to get \(du = -3\,dx\) or \(dx = -\frac{1}{3}\,du\).3. Replace in the integral and simplify:\[\int 3e^{-3x} \, dx = \int -e^{u} \, du = -e^{u} + C\]4. Substitute back to complete the solution: revert \(u = -3x\) giving \(-e^{-3x} + C\).Integration by substitution is especially helpful when the derivative of one part of the function is another portion of the function itself. This lends to neat cancellations, and the integration becomes effortless.