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Chapter 5: Problem 13

Find the average value of each function over the given interval. \(f(x)=3\) on [10,50]

Short Answer

Expert verified
The average value of the function is 3.

Step by step solution

01

Understand the Problem

We need to find the average value of the constant function \(f(x) = 3\) over the interval \([10, 50]\). A constant function means that the function value is the same at every point in the interval.
02

Use the Average Value Formula

The average value of a function \(f(x)\) over an interval \([a, b]\) is given by the formula: \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. Here, \(a = 10\) and \(b = 50\), and \(f(x) = 3\).
03

Set Up the Integral

For the constant function \(f(x) = 3\), the integral becomes \( \int_{10}^{50} 3 \, dx \).
04

Compute the Integral

Calculating the integral of a constant, \( \int 3 \, dx = 3x + C\), over the interval \([10, 50]\), gives \( \left[ 3x \right]_{10}^{50} = 3(50) - 3(10)\).
05

Evaluate the Integral

Calculate \(3(50) - 3(10)\) which results in \(150 - 30 = 120\).
06

Find the Average Value

Substitute back into the average value formula: \[ \text{Average Value} = \frac{1}{50 - 10} \cdot 120 = \frac{1}{40} \cdot 120 = 3 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Function
A constant function is one of the simplest types of functions in mathematics. This function maintains the same output for any input in its domain. When we consider a constant function like \( f(x) = 3 \), it means that no matter what value of \( x \) you plug in, the output will always be 3. This idea makes calculations, such as finding the average or integrating, straightforward.
  • A constant function does not change, which means its graph is a horizontal line.
  • The slope of a constant function is zero, ensuring the function value remains the same at all points.
  • In terms of real-life applications, constant functions can model situations where a value remains steady over time, such as a fixed rate or cost.
Definite Integral
The definite integral is a fundamental concept in calculus that allows us to calculate the accumulated value of a function over a specific interval. For example, when dealing with the function \( f(x) = 3 \), the definite integral computes the total quantity of the constant function over that specified range.
To integrate a constant function like \( f(x) = 3 \) from 10 to 50, you perform the integral calculation:
  • For any constant \( k \) over an interval \([a, b]\), the integral is given by \( \int_{a}^{b} k \, dx = k(b-a) \).
  • This simplifies the task as you multiply the constant by the length of the interval.
The definite integral allows not just for calculating areas under curves but helps in finding more practical solutions, such as average values, as needed in this exercise.
Average Value Formula
The average value of a function over an interval gives us a single value representing the entire function's behavior within that range. It's like boiling down all the values a function takes to one representative value. In mathematical terms, the average value of a function \( f(x) \) over an interval \([a, b]\) is computed using this vital formula:
  • \[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]
  • This formula divides the integral, representing the total accumulation, by the interval width \( b-a \), which balances the function's values over that span.
  • For our specific function \( f(x) = 3 \), the constant nature simplifies the process, making the average value equal to the function's value (since there's no variation).
Utilizing this formula is critical in both pure mathematics and practical scenarios, where understanding an underlying average can illuminate trends or simplify complex analyses.

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Most popular questions from this chapter

Find \(\int(x+1) d x\) : a. By using the formula for \(\int u^{n} d u\) with \(n=1\). b. By dropping the parentheses and integrating directly. c. Can you reconcile the two seemingly different answers? [Hint: Think of the arbitrary constant.]

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int \frac{x}{\sqrt{x+2}} d x $$

A friend says that if you can move numbers across the integral sign, you can do the same for variables since variables stand for numbers, and in this way you can always "fix" the differential \(d u\) to be what you want. Is your friend right?

For each definite integral: a. Evaluate it "by hand." b. Check your answer by using a graphing calculator. $$ \int_{0}^{4} \sqrt{x^{2}+9} x d x $$

BUSINESS: Money Stock Measure From 1964 to 2014 the money stock measure "M1" (currency, traveler's checks, demand deposits, and other checkable deposits) was growing at the rate of approximately \(94 e^{0.56 x}\) billion dollars per decade, where \(x\) is the number of decades since 1964 . Find the total increase in M1 from $$ 1964 \text { to } 2014 $$
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