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Chapter 5: Problem 7

Find the average value of each function over the given interval. \(f(x)=2 x+1\) on [0,4]

Short Answer

Expert verified
The average value of the function is 5.

Step by step solution

01

Understand the Problem

To find the average value of a continuous function \(f(x)\) over a specific interval \([a, b]\), we calculate the integral of the function over the interval and divide by the interval's length.
02

Write the Formula

The formula to find the average value of a function \(f(x)\) on the interval \([a, b]\) is given by:\[\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\]In this case, \(a = 0\), \(b = 4\), and \(f(x) = 2x + 1\).
03

Set Up the Integral

Substitute the given function and limits into the integral:\[\int_{0}^{4} (2x + 1) \, dx\]
04

Calculate the Integral

To evaluate \(\int_{0}^{4} (2x + 1) \, dx\), find the antiderivative:\[ \int (2x + 1) \, dx = x^2 + x + C \]Evaluate from 0 to 4:\[\left[ x^2 + x \right]_{0}^{4} = (4^2 + 4) - (0^2 + 0) = 16 + 4 = 20\]
05

Divide by the Interval Length

The length of the interval \([0, 4]\) is \(4 - 0 = 4\). Now, divide the integral result by this interval length:\[\text{Average value} = \frac{20}{4} = 5\]

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Key Concepts

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

calculus integral
In calculus, the integral is a fundamental concept, representing the area under a curve of a function. The integral calculates the total accumulation of values, which can relate to many real-world phenomena like distance, area, or total income over time.
For a given function, when you integrate over a specific interval [a, b], what you do is essentially summing up every infinitesimally small area under the curve from point a to point b.
  • The process of integration makes use of the integral sign: \[ \int \]
  • It uses limits of integration, such as \( a \) and \( b \) in our example, and expresses the function: \( f(x) \).
The result of this operation often yields another function (often called the anti-derivative) which, when evaluated over the given limits, provides a numerical representation of the area.
antiderivative
The antiderivative, or indefinite integral, is an essential tool that we use to "reverse" differentiate a function. Imagine if you are given the derivative of a function and asked to find the original function鈥攖hat's what finding an antiderivative involves.
  • An antiderivative for a function \( f(x) \) is a function \( F(x) \) whose derivative is \( f(x) \).
  • The notation for the antiderivative includes a constant of integration, \( C \), represented as \[ \int f(x) \, dx = F(x) + C \]
In our example involving the function \( f(x) = 2x + 1 \), calculating the antiderivative by integration gives us \( x^2 + x + C \). Evaluating this antiderivative at the endpoints of the interval helps in finding the definite integral, which we further use to find the average value.
continuous function
A continuous function is one where small changes in input yield small changes in output. You can visualize it as a smooth curve without breaks or holes.
  • Mathematically, a function \( f(x) \) is continuous over an interval if at every point \( c \) in the interval, \( \lim_{x \to c} f(x) = f(c) \).
This attribute of functions is crucial because only continuous functions can be integrated conveniently across an interval.
For example, our function \( f(x) = 2x + 1 \) is continuous over the interval [0, 4], which makes it suitable for applying the integral to find its average value over this interval.
interval analysis
Interval analysis involves breaking down the domain of a function to examine its behavior over a range. The interval [a, b] acts as the window through which we observe the function's properties.
  • In the context of finding average values, this interval determines the limits of integration.
  • The formulation \([a, b]\) signifies we only focus on those x-values between a and b inclusively.
Applying interval analysis helps in seamlessly defining where to start and stop gathering data from a function. For instance, in the exercise, we examine \( f(x) = 2x + 1 \) over the interval [0, 4] to compute its average value. We calculate the integral of \( f(x) \) over this interval and then divide by the interval鈥檚 length (4-0) to get the average. This process provides insight into the function's behavior across the specified domain.

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Most popular questions from this chapter

In 2000 the birthrate in Africa increased from \(17 e^{0.02 t}\) million births per year to \(22 e^{0.02 t}\) million births per year, where \(t\) is the number of years since 2000 . Find the total increase in population that will result from this higher birthrate between \(2000(t=0)\) and \(2050(t=50)\).

For each definite integral: a. Evaluate it "by hand." b. Check your answer by using a graphing calculator. $$ \int_{2}^{3} \frac{x^{2}}{x^{3}-7} d x $$

Suppose that a company found its sales rate (in sales per day) if it did advertise, and also its (lower) sales rate if it did not advertise. If you integrated "upper minus lower" over a month, describe the meaning of the number that you would find.

An average young female in the United States gains weight at the rate of \(14(x-10)^{-1 / 2}\) pounds per year, where \(x\) is her age \((10 \leq x \leq 19)\). Find the total weight gain from age 11 to \(19 .\)

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). $$ \int x \sqrt[3]{x-4} d x $$
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