Problem 17 - Calculus Single Variable

Question

Compute the average value \(f_{\text {avg }}\) of \(f\) over \([a, b]\) and find a point \(c\) in \((a, b)\) for which \(f(c)=f_{\text {avg. }}\) Illustrate the geometric significance of \(c\) with a sketch accompanied by a description. \(f(x)=\sqrt{x} \quad a=1, b=4\)

Step-by-Step Solution

Answer

The average value of \( f \) is \( \frac{14}{9} \), and \( c = \frac{196}{81} \) satisfies \( f(c) = f_{\text{avg}} \). Geometrically, the area under \( f \) from 1 to 4 equals the area of a rectangle with that height.

Step 1: Compute the Average Value of the Function

The average value of a function \( f(x) \) over the interval \([a, b]\) is given by the formula: \[ f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] For \( f(x) = \sqrt{x} \), \( a = 1 \), and \( b = 4 \), we have: \[ f_{\text{avg}} = \frac{1}{4-1} \int_{1}^{4} \sqrt{x} \, dx \] To solve this, compute the integral \( \int \sqrt{x} \, dx \), which is \( \frac{2}{3}x^{3/2} + C \). Next, evaluate the definite integral: \[ \left[ \frac{2}{3}x^{3/2} \right]_{1}^{4} = \frac{2}{3}(4^{3/2}) - \frac{2}{3}(1^{3/2}) \] This simplifies to \( \frac{2}{3}(8) - \frac{2}{3}(1) = \frac{16}{3} - \frac{2}{3} = \frac{14}{3} \). Thus, the average value \( f_{\text{avg}} = \frac{1}{3} \times \frac{14}{3} = \frac{14}{9} \).

Step 2: Find Point c for Which f(c) = f_avg

We need to find \( c \) in \( (1, 4) \) such that \( f(c) = f_{\text{avg}} \). This means \( \sqrt{c} = \frac{14}{9} \). Square both sides to solve for \( c \): \[ c = \left( \frac{14}{9} \right)^2 = \frac{196}{81} \] This value \( c \in (1, 4) \) satisfies \( f(c) = f_{\text{avg}} \).

Step 3: Illustrate the Geometric Significance

The geometric significance of point \( c \) is that it represents a point where the height of the constant function \( f_{\text{avg}} \) over \( [1, 4] \) is equal to the height of \( f(x) = \sqrt{x} \) at \( x = c \). Imagine the graph of \( f(x) = \sqrt{x} \) from \( x = 1 \) to \( x = 4 \). The area under \( f(x) \) is equal to the area of the rectangle with base \( [1, 4] \) and height \( f_{\text{avg}} = \frac{14}{9} \). Thus, \( c \) is where \( f(c) \) equals that constant height, providing a visual balance or "average."

Key Concepts

Integral Calculus

Integral calculus is an essential part of calculus that deals with integrals and their applications. This branch of mathematics helps us find the total accumulation of quantities and the area under curves, among other things. When we say we are solving an integral, we are essentially calculating this total accumulation.
In the original exercise, the integral calculus tool is crucial because it allows us to determine the average value of the function \( f(x) \) over a specific interval \([a, b]\). By using integration, we sum up infinitesimal parts of the area under the curve of \( f(x) \) and then divide by the length of the interval.

• The integral calculus formula involved is \( \int_a^b f(x) \, dx \).
• This provides the total area under \( f(x) \) from \( a \) to \( b \).
• If we plug in \( \sqrt{x} \) for \( f(x) \) and solve, we get the required area under \( f(x) = \sqrt{x} \) from 1 to 4.

By mastering integral calculus, we can tackle a wide range of problems involving areas, accumulated quantities, and other real-world applications. Understanding this concept helps us appreciate how flexible and powerful mathematics can be.

Definite Integral

In integral calculus, a definite integral is used to calculate the exact area under a curve between two points. This area represents either total accumulation or a total change over that interval. The notation \( \int_a^b f(x) \, dx \) indicates that we are finding the definite integral of \( f(x) \) from \( x = a \) to \( x = b \).
In the given exercise, the definite integral is critical for finding the average value of the function \( f(x) = \sqrt{x} \) over the interval \([1, 4]\). The steps involved include:

• Setting up the definite integral \( \int_1^4 \sqrt{x} \, dx \).
• Evaluating this integral: \[ \left[ \frac{2}{3}x^{3/2} \right]_{1}^{4} \]
• Simplifying to get \( \frac{14}{3} \), which represents the area under the curve.

Using the definite integral, we not only find the total area but also help in calculations like average values and other important measures. It gives us a numerical value that quantifies the total effect of \( f(x) \) across the interval \([a, b]\). The definite integral is a core topic in calculus and finds application in physics, engineering, and economics.

Geometric Significance

Geometric significance refers to the visual interpretation of mathematical concepts, which can deepen our understanding. The geometric significance of the average value of a function involves how we perceive this average value on a graph.
In our exercise, the function \( f(x) = \sqrt{x} \) defines a curve over the interval \([1, 4]\). Calculating the average value \( f_{\text{avg}} \) means finding a horizontal line \( y = f_{\text{avg}} \) such that the area under this line (over \([1, 4]\)) equals the area under \( f(x) \) over the same interval.

• Point \( c \) where \( f(c) = f_{\text{avg}} \) marks where \( f(x) \) meets the line at avg height.
• This \( c = \frac{196}{81} \ (\approx 2.42)\) satisfies the equation due to the properties of \( \sqrt{x} \).
• The rectangle's area equals the area under the square root curve, demonstrating a balanced average.

This geometric balance or equality helps us visualize concepts that might otherwise seem abstract. Seeing where \( f(x) \) equals \( f_{\text{avg}} \) provides a meaningful visual representation of averaging and the definite integral insights.