Problem 2 - Calculus for Biology and Medicine

Question

All the integrals are improper and converge. Explain in each case why the integral is improper, andevaluate each integral. $$ \int_{0}^{\infty} x e^{-x} d x $$

Step-by-Step Solution

Answer

The integral converges and evaluates to 1.

Step 1: Identify why the integral is improper

The integral \(\int_{0}^{\infty} x e^{-x} dx\) is improper because the upper limit of integration is infinite. When the range of an integral is infinite, it is classified as improper.

Step 2: Set up the improper integral limit

We convert the improper integral into a limit by replacing the infinite upper bound with a variable and taking the limit as the variable approaches infinity. \[ \int_{0}^{\infty} xe^{-x} dx = \lim_{b \to \infty} \int_{0}^{b} xe^{-x} dx \]

Step 3: Use integration by parts

To evaluate \(\int_{0}^{b} xe^{-x} dx\), use integration by parts. Let \(u = x\) and \(dv = e^{-x} dx\). Then \(du = dx\) and \(v = -e^{-x}\). The formula for integration by parts is \(\int u dv = uv - \int v du\).

Step 4: Apply integration by parts formula

Substitute into the formula: \[ \int_{0}^{b} xe^{-x} dx = \left. -xe^{-x} \right|_{0}^{b} + \int_{0}^{b} e^{-x} dx \]

Step 5: Evaluate the boundary terms and second integral

Evaluate: \[ -xe^{-x} \bigg|_{0}^{b} = [-be^{-b} - 0] \] Next, evaluate \(\int_{0}^{b} e^{-x} dx\): \[ \int e^{-x} dx = -e^{-x} \] Thus, \[ \int_{0}^{b} e^{-x} dx = \left. -e^{-x} \right|_{0}^{b} = [-e^{-b} + 1] \]

Step 6: Combine results and evaluate the limit

Combine and evaluate: \[ -be^{-b} + [-e^{-b} + 1] = -(b+1)e^{-b} + 1 = 1 - (b+1)e^{-b} \] Take the limit as \(b \to \infty\):\[ \lim_{b \to \infty} \left( 1 - (b+1)e^{-b} \right) \] As \(b\) approaches infinity, \((b+1)e^{-b}\) approaches 0, so:\[ \lim_{b \to \infty} \left( 1 - (b+1)e^{-b} \right) = 1 \]

Step 7: Conclude the evaluation

The original improper integral converges to 1. Therefore, the value of the integral \(\int_{0}^{\infty} xe^{-x} dx\) is 1.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique often used to solve integrals involving products of functions. It comes in particularly handy when dealing with exponential functions and polynomials, like in the exercise at hand.

The formula for integration by parts is given by:

• \[ \int u \, dv = uv - \int v \, du \]

In this formula, \(u\) and \(dv\) are parts of the original integral. You choose them based on what will simplify the integral after applying the formula. For the exercise:

• Let \(u = x\) and \(dv = e^{-x} dx\).
• Compute \(du = dx\) and \(v = -e^{-x}\).
• Substitute these into the integration by parts formula.

Proper use of integration by parts helps break down a complex integral into simpler parts, which are then easier to evaluate.

Convergence

Convergence is a key concept when dealing with improper integrals. An integral is said to converge if its value approaches a finite number as the limit goes to infinity, or as any asymptotic bounds are reached. This concept ensures that while infinite scopes are involved in calculation, the result remains quite manageable and finite.

In the given exercise, convergence is tested by examining the limit of the integral as the upper boundary approaches infinity. This integral converges because, as detailed in the solution, the boundary term and exponential decay together yield a finite limit:

• The term \((b+1)e^{-b}\) approaches zero as \(b\) heads to infinity.
• Consequently, the entire expression \(1 - (b+1)e^{-b}\) approaches 1.

Convergence assures us that although we're evaluating over an infinite range, the result is meaningful and finite.

Improper Integrals Evaluation

Improper integrals are those with infinite limits of integration or unbounded integrands. Evaluating these correctly ensures our solutions remain practical and understandable. The integrals often require converting the improper integral into a limit to facilitate evaluation.

In this exercise, the improper nature was due to the infinite limit of integration. Hence, to evaluate:

• Replace the infinite limit with a variable, typically \(b\), and compute the integral with respect to this variable.
• Take the limit as this variable approaches infinity: \[ \int_{0}^{\infty} xe^{-x} dx = \lim_{b \to \infty} \int_{0}^{b} xe^{-x} dx \]
• Evaluate the defined integral, and use limits to find its finite sum.

Calculating improper integrals in this manner allows for integrating functions across limitless domains while reaching finite results. It's an essential tool in calculus, well used in analyzing scientific and engineering problems involving infinite or undefined scales.