Problem 2 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The integral is improper due to an infinite upper limit and evaluates to 1.

Step 1: Identify the Improper Integral

The integral \( \int_{0}^{\infty} x e^{-x} \, dx \) is improper because its upper limit of integration is \( \infty \). This requires us to take a limit as \( b \to \infty \), transforming the integral into a proper integral.

Step 2: Transform the Improper Integral into a Limit

Express the improper integral with a limit: \[ \int_{0}^{\infty} x e^{-x} \, dx = \lim_{b \to \infty} \int_{0}^{b} x e^{-x} \, dx \] This allows us to work with finite bounds before taking the limit.

Step 3: Perform Integration by Parts

Use integration by parts, where \( u = x \) and \( dv = e^{-x} \, dx \). Then \( du = dx \) and \( v = -e^{-x} \). Apply the formula \( \int u \, dv = uv - \int v \, du \): \[ \int_{0}^{b} x e^{-x} \, dx = \left[ -x e^{-x} \right]_{0}^{b} + \int_{0}^{b} e^{-x} \, dx \]

Step 4: Evaluate Each Component

Evaluate the first component: \( \left[ -x e^{-x} \right]_{0}^{b} = -b e^{-b} + (0 \cdot e^{0}) = -b e^{-b} + 0 \). Evaluate the second integral: \( \int_{0}^{b} e^{-x} \, dx = \left[ -e^{-x} \right]_{0}^{b} = -e^{-b} + 1 \).

Step 5: Combine the Results

Combine the results of integration by parts: \[ \lim_{b \to \infty} \left( -b e^{-b} - e^{-b} + 1 \right) \]To find the limit as \( b \to \infty \), note that both \( b e^{-b} \to 0 \) and \( e^{-b} \to 0 \). Therefore, the limit becomes: \[ 0 + 0 + 1 = 1 \]

Step 6: Conclusion

The evaluation shows that \( \int_{0}^{\infty} x e^{-x} \, dx = 1 \). This indicates that while the integral is improper due to the infinite upper limit, it converges to a value of 1.

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique in calculus used to transform the integral of a product of functions into simpler terms. It's crucial when facing integrals that involve products of polynomial and exponential functions, like the one given: \[ \int x e^{-x} \, dx \] This method relies on the formula: \[ \int u \, dv = uv - \int v \, du \] Applying this formula, you first choose parts of the integrand as "\(u\)" and "\(dv\)":

• Let \( u = x \) and \( dv = e^{-x} \, dx \).
• This gives \( du = dx \) and \( v = -e^{-x} \).

Substituting these into the integration by parts formula allows the integral to simplify into more manageable components:

• \[ \int x e^{-x} \, dx = \left[ -x e^{-x} \right] + \int e^{-x} \, dx \]
• The first term evaluates directly within specified bounds, and the second integral is straightforward.

Together, these evaluations lead directly to the required convergence step, demonstrating how integration by parts assists in breaking down complex integrals.

Convergence of Integrals

Improper integrals, such as \( \int_{0}^{\infty} x e^{-x} \, dx \), require special attention to understand if they converge to a finite value. In our case, the challenge is the infinity at the upper limit of the integral. To assess convergence, a common method is to replace the infinite limit with a finite variable and evaluate: \[ \lim_{b \to \infty} \int_{0}^{b} x e^{-x} \, dx \]This approach helps to circumvent the undefined nature posed by infinity.
It's crucial to compute whether the limit provides a sensible number. For the given function, after performing integration and evaluating the limit, the result of convergence is a finite number, 1. This indicates the area under the curve from zero to infinity is finite and well-defined, showcasing convergence.

Limits in Calculus

Limits serve as the backbone of calculus and are essential when dealing with improper integrals. In this problem, the function \( \int_{0}^{b} x e^{-x} \, dx \) becomes finite by introducing a limit as \( b \) approaches infinity.
Evaluating this limit requires understanding that expressions like \( b e^{-b} \) and \( e^{-b} \) both approach zero as \( b \) grows huge.

• \( b e^{-b} \to 0 \) because the exponential decay of \( e^{-b} \) outpaces the linear growth of \( b \).
• \( e^{-b} \to 0 \) since any exponent at infinity shrinks non-zero bases to near zero values.

Adding these results leads to the limit expressing the integral's evaluation as 1. The role of limits reveals how calculus manages otherwise undefined ranges, affirming that the initial integral converges.