Problem 43 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the average value of the function f over the indicated interval \([a, b]\). $$f(x)=x e^{x^{2}} ;[0,2]$$

Step-by-Step Solution

Answer

The average value of the function \(f(x) = xe^{x^2}\) over the interval \([0, 2]\) is \(\frac{1}{4} (e^4 - 1)\).

Step 1: Find the integral of the function

First, we need to find the integral of the function \(f(x) = xe^{x^2}\) over the given interval \([0, 2]\). To do this, we will set up the integral: $$\int_{0}^{2} f(x) dx = \int_{0}^{2} xe^{x^2} dx$$ To evaluate this integral, we use substitution method. Let \(u = x^2\), then \(\frac{du}{dx} = 2x\), so \(\frac{1}{2} du = x \, dx\). Now, we substitute \(u\) into the integral and also change the limits of integration accordingly: when \(x = 0\), \(u = 0^2 = 0\); when \(x = 2\), \(u = 2^2 = 4\). Now we have: $$\int_{0}^{2} xe^{x^2} dx = \int_{0}^{4} e^u \frac{1}{2} du$$ Now we can evaluate the integral: $$\int_{0}^{4} e^u \frac{1}{2} du = \frac{1}{2} \int_{0}^{4} e^u du$$ $$= \frac{1}{2} [e^u]_{0}^{4} = \frac{1}{2} (e^4 - e^0)$$

Step 2: Divide the result by the length of the interval

Now we need to divide this result by the length of the interval to find the average value. The length of the interval \([0, 2]\) is \(2 - 0 = 2\). So, we get: $$\text{Average value of} \, f(x) = \frac{\int_{0}^{2} f(x) dx}{2} = \frac{\frac{1}{2} (e^4 - e^0)}{2}$$

Step 3: Simplify the result

Finally, we simplify the result to get the average value: $$\text{Average value of} \, f(x) = \frac{\frac{1}{2} (e^4 - e^0)}{2} = \frac{1}{4} (e^4 - 1)$$ So, the average value of the function \(f(x) = xe^{x^2}\) over the interval \([0, 2]\) is \(\frac{1}{4} (e^4 - 1)\).

Key Concepts

Average Value of a Function

The average value of a function over an interval gives us an understanding of how the function behaves in a particular range. Imagine you're trying to determine a typical or mean value for the function across that interval, similar to finding an average grade in a class.
To compute this, you integrate the function over the specified interval and then divide by the length of that interval. It's like taking a sum and then finding its mean.

• Formula: If you have a function \( f(x) \) on an interval \( [a, b] \), its average value is calculated as:\[\text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]
• For example, for the function \( f(x) = xe^{x^2} \) over \([0, 2]\), the task involves integrating and then dividing by \(2 - 0 = 2\).

Understanding this concept helps to appreciate the overall behavior of the function over the interval, rather than focusing on just a single point.

Definite Integral

The definite integral is a fundamental concept in calculus that calculates the accumulation of quantities. Geometrically, it represents the area under the curve of a function, between two points on the x-axis. It's used to add up infinite small slices of the area under the curve.

• A definite integral is written as \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) define the interval of integration.
• For our example, the definite integral \( \int_{0}^{2} xe^{x^2} \, dx \) represents the area between the function and the x-axis from x = 0 to x = 2.

The result of this process is not just a raw value, but can give insights into how a given quantity behaves over a specific range.

Integration by Substitution

Integration by substitution is a technique similar to the substitution method in algebra, where you replace a part of your equation with a simpler equivalent. The goal is to simplify the integral, making it easier to solve.
This method involves:

• Choosing a substitution, typically \( u = g(x) \), that simplifies \( f(x) \, dx \) to \( f(g(x))g'(x) \, dx \).
• Replacing \( g(x) \) with \( u \), and adjusting the differential and limits if it’s a definite integral.

In the exercise, \( u \) was set as \( x^2 \), simplifying the process to resolve the integral from 0 to 4, instead of the original from 0 to 2.

Interval of Integration

The interval of integration refers to the range over which the integration occurs. Think of it as the starting and ending points of your travel along the x-axis when calculating the area under the curve.

• The interval is denoted as \([a, b]\), with \(a\) being the lower bound and \(b\) the upper bound.
• For the function \( f(x) = xe^{x^2} \), the specified interval was \([0, 2]\).

Changing the interval can significantly impact the result of the integral, much like changing the start or end of a race will affect the total distance traveled. Understanding and selecting the correct interval is crucial for accurate integration.