Problem 15 - Biocalculus Calculus for the Life Sciences

Question

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \(\int_{0}^{\infty} s e^{-5 s} d s\)

Step-by-Step Solution

Answer

The integral is convergent and its value is \( \frac{1}{25} \).

Step 1: Analyze the Type of Integral

The given integral is \[ \int_{0}^{\infty} s e^{-5s} \, ds. \]This is an improper integral because it has an infinite upper limit. We need to determine if it converges or diverges and find the integral's value if it converges.

Step 2: Set Up the Limit for Evaluation

To handle the improper integral, we express it as a limit:\[ \int_{0}^{\infty} s e^{-5s} \, ds = \lim_{b \to \infty} \int_{0}^{b} s e^{-5s} \, ds. \]

Step 3: Use Integration by Parts

Use integration by parts where you choose:- \( u = s \rightarrow du = ds \)- \( dv = e^{-5s} \, ds \rightarrow v = \int e^{-5s} \, ds = -\frac{1}{5} e^{-5s} \)Apply the integration by parts formula:\[ \int u \, dv = uv - \int v \, du. \]This gives us:\[ \int_{0}^{b} s e^{-5s} \, ds = \left. -\frac{1}{5} s e^{-5s} \right|_{0}^{b} + \frac{1}{5} \int_{0}^{b} e^{-5s} \, ds. \]

Step 4: Evaluate the Parts

Calculate the expression:- The first part is\[ -\frac{b}{5} e^{-5b} - \left(-\frac{0}{5} \times e^{0}\right) = -\frac{b}{5} e^{-5b}. \]- For the second integral:\[ \frac{1}{5} \int_{0}^{b} e^{-5s} \, ds = \frac{1}{5} \left[-\frac{1}{5} e^{-5s}\right]_{0}^{b} = \frac{1}{25} \left(e^{-5 \times 0} - e^{-5b}\right) = \frac{1}{25} (1 - e^{-5b}). \]

Step 5: Combine and Evaluate the Limit

Combine the evaluated parts to form:\[ -\frac{b}{5} e^{-5b} + \frac{1}{25} (1 - e^{-5b}). \]Take the limit as \( b \to \infty \):- As \( b \to \infty \), \( e^{-5b} \to 0 \), thus\[ -\frac{b}{5} \times 0 = 0, \]allowing us to simplify the second term to\[ \lim_{b \to \infty} \frac{1}{25} (1 - 0) = \frac{1}{25}. \]

Step 6: Conclusion

The evaluated limit \( \frac{1}{25} \) is a finite number, indicating that the integral converges and its value is \( \frac{1}{25} \).

Key Concepts

Improper Integrals

Improper integrals often appear when you encounter infinite intervals or unbounded functions. In the exercise, the integral \[ \int_{0}^{\infty} s e^{-5s} \, ds\] is classified as improper due to its upper limit, infinity. Dealing with improper integrals means understanding how their behavior influences the convergence or divergence. Here's why it matters:

• Convergence: When an improper integral converges, it means the area under the curve approaches a specific value.
• Divergence: Conversely, divergence indicates that the area does not approach a finite value.

Handling these requires transforming the original integral with limits to inspect its behavior. Always remember to look for conditions like infinite limits or discontinuous functions as these are key indicators of an improper integral.

Integration by Parts

Integration by parts is a vital tool for calculating integrals that are products of functions, and it echoes the product rule for differentiation. The formula is expressed as:\[ \int u \, dv = uv - \int v \, du.\]
For our exercise, selections were made as follows:

• Letting \( u = s \) and \( dv = e^{-5s} \, ds \).
• Then, differentiating and integrating these gives \( du = ds \) and \( v = -\frac{1}{5}e^{-5s} \).

The substitution allows us to resolve complex integrals into simpler ones. This transformation reduces the complexity of the integral by breaking it into parts that are easier to handle. It's especially useful when one part of the function becomes simpler through differentiation or integration.

Limits and Convergence

Crucial to solving improper integrals is understanding limits and convergence. For the integral to converges means that it approaches a finite value as the bound approaches infinity. Here's the breakdown:

• Expressing the integral as a limit allows for evaluation over an infinite interval: \[ \int_{0}^{\infty} s e^{-5s} \, ds = \lim_{b \to \infty} \int_{0}^{b} s e^{-5s} \, ds.\]
• After applying integration by parts, you evaluate each component as \( b \to \infty \).
• In this case, simplifying to a cleaner form makes it easy to see if as \( b \to \infty \), terms tend to zero, indicating convergence.

In practice, evaluate the behavior of each term in these contexts to understand the overall convergence or divergence. This approach combines analysis of individual parts with their behavior at extreme bounds.

Calculus in Life Sciences

The concepts of improper integrals are not just abstract mathematical ideas but find numerous applications in life sciences. In biological or pharmacokinetics studies, such integrals can be used to calculate the total amount of a drug in the bloodstream over an infinite period, essentially until fully metabolized.

• For example, the integral could represent the quantity of a radioactive substance that decays over time.
• Protein folding probabilities, population growth models, and diffusion processes often require solving improper integrals to determine steady-states or cumulative dynamic behaviors.

Understanding how functions behave over time is a crucial element when modeling biological processes. Calculus provides the tools to sum up infinite processes, capturing the essence of biological realities in mathematical terms. This is essential for developing simulations and algorithms that mirror real-life biological systems.