91Ó°ÊÓ

Calculating the average length of a molecule is a common task in fields such as chemistry and biology. The average length is often represented by \( \bar{x} \) and is determined using calculus and integrals. In this particular scenario, we use an integral to compute the average molecular length, where the function integrates over a mathematical model of the molecule’s length distribution.

In this exercise, the average length \( \bar{x} \) is expressed as:

• \( \bar{x} = \lim_{b \rightarrow \infty} \left[ 0.1 \int_{0}^{b} x^3 e^{-x^2 / 8} \, dx \right] \)

This integral considers a distribution where \( x^3 e^{-x^2/8} \) models how the likelihood of a molecule having a certain length \( x \) decays exponentially.

We are tasked with evaluating the integral and then determining the limit as \( b \) approaches infinity. By performing the calculations, we approximate that \( \bar{x} \) is roughly equal to 3.2 nm, indicating the average molecular length.

Integrating functions that include an exponential term is a common but important task in mathematics. An integral containing an exponential function, such as \( e^{-x^2/8} \), can sometimes be tricky because exponential functions have properties that influence the behavior of the entire integral.

To simplify the process, we often use substitution. In this scenario, we performed a substitution where \( u = \frac{x^2}{8} \). The idea is to reframe the integrals in terms of \( u \) rather than \( x \), which can simplify the mathematical operations required to solve it.

After changing the integration limits to accommodate this substitution, the original integral transformed into:

• \( \int_{0}^{b^2/8} 32u^{3/2} e^{-u} \, du \)

Here, the substitution helps manage the exponential behavior and allows us to apply integration techniques like integration by parts or reference to integrals of a known form.

Evaluating limits is crucial when trying to understand behavior as variables approach a large value or infinity. Limits help determine how functions behave and converge, providing vital insights in calculus.

In our example, we evaluate the limit as \( b \rightarrow \infty \) for the integral:

• \( \bar{x} = \lim_{b \rightarrow \infty} \left[ 0.1 \int_{0}^{b} x^3 e^{-x^2/8} \, dx \right] \)

The behavior of the integral at infinity is dominated by the exponential term \( e^{-x^2/8} \), which causes the integral to stabilize to a finite value. This occurs because exponential decay rapidly reduces the contributions from increasingly large \( x \) values.

The calculation confirms that, as \( b \) approaches infinity, the limit yields a finite value, approximately 3.2 nm. The limit helpfully truncates any unbounded growth or undefined behavior, providing us with a meaningful result.

Improper integrals are used when limits of integration include infinity or when the function has undefined behavior at certain points within the integration bounds. Integrals extending to infinity or approaching undefined points require special attention to ensure they converge to a finite value.

In this exercise, we evaluate an improper integral with an upper limit extending towards infinity:

• \( \int_{0}^{b} x^3 e^{-x^2/8} \, dx \)

The upper bound \( b \to \infty \) requires evaluating the behavior of the function across its entire range. An efficient method to tackle these integrals involves utilizing limits, recognizing factors like quick decay rates as significant contributors to convergence.

These improper integrals rely on the balancing act between exponential decay and polynomial growth to ensure a finite result, demonstrating the versatility and power of integration in extracting meaningful, quantitative insights from seemingly divergent expressions.