Calculating the Average Value of the Function f(x) (1 - ln(x))/x over the Interval [1, e]

In this article, we explore the process of finding the average value of the function f(x) (1 - ln(x))/x over the interval [1, e]. This involves evaluating the integral of the function and applying the formula for the average value of a function over a given interval.

Introduction to the Problem

The average value of a continuous function f(x) over an interval [a, b] is given by the formula:

A 1/(b - a) ∫ab f(x) dx

For our case, the interval is [1, e], and the function is f(x) (1 - ln(x))/x. Thus, the average value A is:

A 1/(e - 1) ∫1e (1 - ln(x))/x dx

Solving the Integral

To evaluate this integral, we start with the given integral:

A 1/(e - 1) ∫13 (1 - ln(x))/x dx

However, the interval [1, e] suggests we should change the bounds to match [1, e].

Substitution Method

Let's use the substitution method with x ey. This implies that dx ey dy. When x 1, then y 0, and when x e, then y 1. Hence, the integral becomes:

A 1/(e - 1) ∫01 (1 - y) e-y dy

Evaluating the Integral

Now, we can evaluate the integral ∫01 (1 - y) e-y dy by splitting it into two parts:

∫01 (1 - y) e-y dy ∫01 e-y dy - ∫01 y e-y dy

Evaluating the first part:

∫01 e-y dy [-e-y]01 1 - e-1

Evaluating the second part using integration by parts:

Let u y, dv e-y dy, then du dy, v -e-y.

∫01 y e-y dy [-y e-y]01 ∫01 e-y dy

[-y e-y]01 ∫01 e-y dy (1 - 1)e-1 C - [e-y]01 -e-1 1 e-1 1 - e-1

Thus,

1 - 1 e-1 e-1

Final Calculation

Now, we combine the results:

A 1/(e - 1) [1 - 1/2] 1/(e - 1) * 1/2 1/2e - 1

Conclusion

The average value of the function f(x) (1 - ln(x))/x over the interval [1, e] is 1/2e - 1. This demonstrates the process of integrating a function and then applying the formula for the average value of a function over a specific interval.

Understanding how to solve such integrals and apply the formula for average values is crucial for various applications in mathematics and real-world scenarios, especially in fields like physics and engineering.