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Chapter 4: Problem 79

Find the average value \(f_{\text {av }}\) of the function over the indicated interval. $$ f(x)=\frac{x}{\sqrt{2} ;} ;[0,3] $$

Short Answer

Expert verified
The average value of the function \(f(x) = \frac{x}{\sqrt{2}}\) over the interval \([0, 3]\) is \(f_{av} = \frac{3}{2\sqrt{2}}\).

Step by step solution

01

Write down Average Value formula

The average value of the function is given by the formula: $$ f_{av} = \frac{1}{b - a}\int_{a}^{b} f(x)dx $$ In our case, \(a=0\), \(b=3\) and \(f(x)=\frac{x}{\sqrt{2}}\).
02

Substitute values into the Average Value formula

Substitute the given values into the formula: $$ f_{av} = \frac{1}{3 - 0}\int_{0}^{3} \frac{x}{\sqrt{2}} dx $$
03

Calculate the integral

Calculate the integral of the function \(\frac{x}{\sqrt{2}}\) from 0 to 3: $$ \int_{0}^{3} \frac{x}{\sqrt{2}} dx = \frac{1}{\sqrt{2}}\int_{0}^{3} x dx $$ Now integrate x: $$ \frac{1}{\sqrt{2}}\left[\frac{1}{2}x^2\right]_{0}^{3} = \frac{1}{\sqrt{2}}\left[\frac{1}{2}(3)^2 - \frac{1}{2}(0)^2\right] $$
04

Simplify the expression

Simplify the expression to find the integral value: $$ \frac{1}{\sqrt{2}}\left[\frac{1}{2}(9)\right] = \frac{9}{2\sqrt{2}} $$
05

Divide integral by the width of the interval

Now divide the integral value by the width of the interval, which is 3: $$ f_{av} = \frac{1}{3}\cdot\frac{9}{2\sqrt{2}} = \frac{9}{6\sqrt{2}} $$
06

Simplify the expression and write the answer

Simplify the expression to obtain the average value of the function over the interval \([0, 3]\): $$ f_{av} = \frac{3}{2\sqrt{2}} $$ So, the average value of the function \(f(x) = \frac{x}{\sqrt{2}}\) over the interval \([0, 3]\) is \(\frac{3}{2\sqrt{2}}\).

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Key Concepts

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

Average Value of Function
In Calculus, the average value of a function on a closed interval \[a, b\], helps us understand the "central tendency" of the function over that interval. It's like finding the arithmetic mean, but for a continuous function.
To calculate the average value of a function \(f(x)\) over an interval \[a, b\], we use this formula:
  • \( f_{av} = \frac{1}{b - a}\int_{a}^{b} f(x)\,dx \)
This tells us that:
  • The average value \(f_{av}\) is the integral of the function over the interval, divided by the length of the interval \((b - a)\).
By integrating the function, we find the total area under the curve, and dividing that by the interval length gives us the average height. This concept forms a foundational idea in understanding how functions behave over specific intervals.
Definite Integral
The definite integral is a major concept in Calculus which helps in finding the net area under a curve defined by a function \(f(x)\) from one point to another, usually known as \[a, b\].
It is represented as:
  • \(\int_{a}^{b} f(x)\,dx\)
The definite integral provides:
  • The total area taking into account areas above and below the x-axis, signifying positive and negative values.
  • Helps in determining total accumulation of quantities, and in the context of this problem, helps to find the overall 'effect' of a function over an interval.
Think of it as adding up all the tiny areas under the curve to get a total area, which can be positive or negative depending on the position relative to the x-axis.
Function Integration
Function integration is the process of finding the integral of a function, and it's crucial for calculating areas under curves and solving differential equations. This involves finding a new function whose derivative matches the original function, known as the anti-derivative.
In the problem at hand:
  • We integrate \(\frac{x}{\sqrt{2}}\) over the interval from 0 to 3.
The iterative steps include finding the antiderivative and evaluating it at the bounds of the interval. This results in:
  • \(\int_{0}^{3} \frac{x}{\sqrt{2}} \,dx = \frac{1}{\sqrt{2}}\left[\frac{1}{2}x^2\right]_{0}^{3}\)
  • This technique is part of a larger technique called the Fundamental Theorem of Calculus, which links the operation of differentiation to integration.
Understanding integration can turn complex problems into manageable analyses, such as calculating net changes or average values over intervals.
Mathematical Formula
Mathematical formulas serve as the cornerstone for problem-solving in Calculus and beyond. They provide a structured approach to solving complex problems and often reveal deep insights about relationships between variables.
In this exercise, the formula for the average value of a function is applied:
  • \( f_{av} = \frac{1}{b - a}\int_{a}^{b} f(x) \,dx \) simplifies the process of finding average values over an interval.
Using mathematical formulas often involves:
  • Substituting known values to move towards a solution (like plugging in \(a=0\), \(b=3\), and the function \(f(x)=\frac{x}{\sqrt{2}}\)).
  • Simplifying expressions and calculations to get the final answer.
Mastering these formulas provides a reliable toolset that enables you to solve a wide variety of mathematical problems more efficiently.

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Most popular questions from this chapter

Velocity of a Sports Car The velocity function for a sports car traveling on a straight road is given by $$ v(t)=\frac{80 t^{3}}{t^{3}+100} \quad 0 \leq t \leq 16$$ where \(t\) is measured in seconds and \(v(t)\) in feet per second. Use Simpson's Rule with \(n=8\) to estimate the average velocity of the car over the time interval \([0,16]\).

Prove that \(\sum_{k=1}^{n}\left(a_{k}-b_{k}\right)=\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} b_{k}\)

a. If \(f\) is even, what can you say about \(\int_{-\pi}^{\pi} f(x) \cos n x d x\) and \(\int_{-\pi}^{\pi} f(x) \sin n x d x\) if \(n\) is an integer? Explain. b. If \(f\) is odd, what can you say about \(\int_{-\pi}^{\pi} f(x) \cos n x d x\) and \(\int_{-\pi}^{\pi} f(x) \sin n x d x ?\) Explain.

Show that $$ \int_{-1}^{1} \sqrt{x^{2}+1} \sec x d x=2 \int_{0}^{1} \sqrt{x^{2}+1} \sec x d x $$

Velocity of an Attack Submarine The following data give the velocity of an attack submarine taken at 10 -min intervals during a submerged trial run. $$\begin{array}{|l|ccccccc|} \hline \text { Time } t \text { (hr) } & 0 & \frac{1}{6} & \frac{1}{3} & \frac{1}{2} & \frac{2}{3} & \frac{5}{6} & 1 \\ \hline \text { Velocity } v \text { (mph) } & 14.2 & 24.3 & 40.2 & 45.0 & 38.5 & 27.6 & 12.8 \\ \hline \end{array}$$ Use Simpson's Rule to estimate the distance traveled by the submarine during the 1 -hr submerged trial run.
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