Problem 41 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

AVERAGE VALUE OF A FUNCTION In Exercises 39 through 42, find the average value of the given function over the indicated interval. \(g(v)=v e^{-v^{2}} ;\) over \(0 \leq v \leq 2\)

Step-by-Step Solution

Answer

\(\frac{1 - e^{-4}}{4}\)

Step 1 - Write down the formula for average value

The formula for the average value of a function over an interval \[a, b\] is given by: \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]

Step 2 - Substitute the given values into the formula

In this exercise, the function is \[ g(v) = v e^{-v^2} \] and the interval is \[ 0 \leq v \leq 2 \. \] Substitute these into the formula: \[ f_{avg} = \frac{1}{2-0} \int_{0}^{2} v e^{-v^2} \, dv. \]

Step 3 - Simplify the expression

Simplify the average value formula: \[ f_{avg} = \frac{1}{2} \int_{0}^{2} v e^{-v^2} \, dv. \]

Step 4 - Integrate the function

To integrate \[ v e^{-v^2} \], use a substitution. Let \[ u = v^2 \] which implies \[ du = 2v \, dv \] or \[ dv = \frac{du}{2v}. \] Substituting these into the integral gives: \[ \int v e^{-v^2} \, dv = \frac{1}{2} \int e^{-u} \, du. \]

Step 5 - Evaluate the integral

Evaluate the integral: \[ \frac{1}{2} \int e^{-u} \, du = \frac{1}{2} [-e^{-u}] = \frac{-e^{-u}}{2}. \]

Step 6 - Change back to variable v and evaluate between the limits 0 and 2

Change back to variable \[ v \] and evaluate the definite integral from \[ 0 \leq v \leq 2 \]: \[ \frac{-e^{-v^2}}{2} \Bigg|_{0}^{2} = \frac{-e^{-4} + e^{0}}{2}\bigg]= \frac{1-e^{-4}}{2}. \]

Step 7 - Find the average value

The average value is: \[ f_{avg} = \frac{1}{2} \left(\frac{1-e^{-4}}{2}\right) = \frac{1-e^{-4}}{4}. \]

Key Concepts

Integration by Substitution

Integration by substitution is a method to simplify an integral by changing variables. This technique is particularly useful when dealing with composite functions. For instance, in the given problem, we need to integrate the function \( v e^{-v^2} \).

To make it simpler, we use a substitution: let \( u = v^2 \). Then, the differential \( du = 2v \, dv \) of this substitution implies \( dv = \frac{du}{2v} \).

By substituting these values into the integral, we change it from an expression in terms of \( v \) into an easier one in terms of \( u \). This transforms the integral to \( \frac{1}{2} \int e^{-u} \, du \), which is much simpler to solve.

Definite Integral

A definite integral calculates the accumulation of a function's values between two limits. This means you're finding the area under the curve of the function between specified bounds. In our problem, we seek the integral of \( g(v) = v e^{-v^2} \) from \( 0 \) to \( 2 \). The formula for a definite integral is: \[ \int_{a}^{b} f(x) \, dx \] where \( a \) and \( b \) are the lower and upper limits, respectively.

After performing the integration, we need to evaluate it at the upper and lower limits, then subtract. For the integral \( \frac{1}{2} \int e^{-u} \, du \), evaluated from \( v = 0 \) to \( v = 2 \), we substitute back and find the area under the curve between these points, which is \( \frac{-e^{-v^2}}{2} \bigg|_{0}^{2} \).

Average Value Formula

The average value of a function over a given interval helps determine the function's central tendency across that interval. The formula for the average value of a function \( f(x) \) over the interval \[ [a, b] \] is given by: \[ f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]

In our problem, the given interval is \( 0 \) to \( 2 \) and the function is \( g(v) = v e^{-v^2} \). Substituting these values into the formula, we get: \[ f_{avg} = \frac{1}{2-0} \int_{0}^{2} v e^{-v^2} \, dv \]. After solving the integral and simplifying, the final average value is: \[ f_{avg} = \frac{1}{2} \left( \frac{1 - e^{-4}}{2} \right) = \frac{1 - e^{-4}}{4}. \]