Problem 15 - Calculus Single Variable

Question

In each of Exercises \(13-16,\) a function \(f\) and an interval \(I=[a, b]\) are given. Find a number \(c\) in \((a, b)\) for which \(f(c)\) is the average value of \(f\) on \(I\). $$ f(x)=x^{2}-10 x / 3 \quad I=[1,3] $$

Step-by-Step Solution

Answer

The value of \( c \) that makes \( f(c) \) equal to the average value on \([1, 3]\) is \( c = \frac{7}{3} \).

Step 1: Understand the Average Value Formula

Recall that the average value of a function \( f \) on the interval \([a, b]\) is given by the formula \[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx. \] This formula provides the mean value of the function over the specified interval.

Step 2: Identify Function and Interval

In this exercise, the function given is \( f(x) = x^2 - \frac{10x}{3} \) and the interval is \([1, 3]\). This means \(a = 1\) and \(b = 3\). We will use these to find the average value.

Step 3: Calculate the Integral

Compute the definite integral of \( f(x) = x^2 - \frac{10x}{3} \) from 1 to 3:\[ \int_{1}^{3} \left( x^2 - \frac{10x}{3} \right) \, dx. \]First, find the antiderivative:- The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \).- The antiderivative of \( -\frac{10x}{3} \) is \( -\frac{10x^2}{6} = -\frac{5x^2}{3} \).Thus, the antiderivative of \( f(x) \) is \( \frac{x^3}{3} - \frac{5x^2}{3} \).

Step 4: Evaluate the Definite Integral

Evaluate \[ \left[ \frac{x^3}{3} - \frac{5x^2}{3} \right]_{1}^{3}. \]Calculate at the upper limit \( x = 3 \): \[ \frac{3^3}{3} - \frac{5 \cdot 3^2}{3} = \frac{27}{3} - \frac{45}{3} = 9 - 15 = -6. \]Calculate at the lower limit \( x = 1 \): \[ \frac{1^3}{3} - \frac{5 \cdot 1^2}{3} = \frac{1}{3} - \frac{5}{3} = -\frac{4}{3}. \]Subtract the lower limit from the upper limit: \[ -6 - ( -\frac{4}{3}) = -6 + \frac{4}{3} = -\frac{18}{3} + \frac{4}{3} = -\frac{14}{3}. \]

Step 5: Calculate the Average Value

Using the integral value found above, compute the average value of \( f \) on \([1, 3]\):\[ \text{Average value} = \frac{1}{3-1} \cdot \left(-\frac{14}{3}\right) = \frac{-14}{6} = -\frac{7}{3}. \]

Step 6: Find the Value \( c \)

We need to find a value \( c \in (1, 3) \) such that \( f(c) = -\frac{7}{3} \). Set the function equal to average:\[ c^2 - \frac{10c}{3} = -\frac{7}{3} \]Multiply through by 3 to clear fractions:\[ 3c^2 - 10c = -7 \]Rearrange into standard form:\[ 3c^2 - 10c + 7 = 0. \]

Step 7: Solve the Quadratic Equation

Solve \( 3c^2 - 10c + 7 = 0 \) using the quadratic formula where \( a = 3 \), \( b = -10 \), \( c = 7 \):\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 7}}{2 \cdot 3} \]\[ c = \frac{10 \pm \sqrt{100 - 84}}{6} = \frac{10 \pm \sqrt{16}}{6} = \frac{10 \pm 4}{6}. \]This gives solutions \( c = \frac{14}{6} = \frac{7}{3} \) and \( c = 1 \). Only \( \frac{7}{3} \in (1, 3) \), so \( c = \frac{7}{3} \).

Key Concepts

Average value of a function

The average value of a function over a specific interval provides insight into the general behavior of the function in that range. This is similar to finding the average of a set of numbers, allowing us to understand the "typical" value of the function across the interval.
The formula to calculate the average value of a continuous function \( f(x) \) from \( a \) to \( b \) is:

• \( \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

This approach integrates the function to capture all the area under the curve over the interval and then divides by the interval length \( b-a \). This division transforms the area into an average height, providing a comprehensive view of the function's behavior.
In practical terms, this concept is beneficial for fields like physics and engineering, where understanding the overall trend over a domain is crucial.

Definite integral

A definite integral enables the calculation of the total area under a curve within a specific interval. This interval is defined by two points \( a \) and \( b \), forming the limits of integration. It represents the aggregation of the function's values across that range, essentially summing the effect of small incremental changes.
The notation \( \int_{a}^{b} f(x) \, dx \) refers to the definite integral of \( f(x) \) from \( a \) to \( b \). It is a fundamental tool in calculus that links the local properties of a function with its global behavior. The process involves finding an antiderivative, substituting the boundary values, and then calculating the net difference to determine the area.
Understanding definite integrals is pivotal for solving real-world problems where cumulative changes need consideration, such as computing total distances or quantities in science and engineering applications.

Quadratic equation

A quadratic equation, composed of a polynomial with the highest degree of two, is expressed in the form \( ax^2 + bx + c = 0 \). Solving a quadratic equation involves finding the values of \( x \) that satisfy this equation.

• The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is a widely used method to solve such equations.

Here, the terms \( a \), \( b \), and \( c \) are coefficients where \( a eq 0 \). The expression under the square root sign \( b^2 - 4ac \) is called the discriminant. It determines the nature of the roots:

• If positive, there are two real and distinct roots.
• If zero, the equation has exactly one real root.
• If negative, the equation has two complex roots.

These characteristics are crucial in predicting the positioning and nature of solutions, providing flexibility for tackling algebraic challenges in various contexts.

Antiderivatives

Antiderivatives, often called indefinite integrals, reverse the process of differentiation. They play a key role in computing definite integrals and solving differential equations.
An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that \( F'(x) = f(x) \). The general form of an antiderivative includes a constant of integration, commonly denoted by \( C \), representing an arbitrary constant that arises because differentiation of a constant is zero.
To solve an integral, one may find the antiderivative of the function being integrated. For example, the antiderivative of \( x^n \) (where \( n eq -1 \)) is \( \frac{x^{n+1}}{n+1} \) plus a constant \( C \).
Antiderivatives are essential not only in integration but also in modeling situations where a function represents a rate of change, providing a foundation for more advanced analyses in calculus.