/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Find the average value of each f... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the average value of each function over the given interval. \(f(x)=x^{n}\) on \([0,1],\) where \(n\) is a constant \((n>0)\)

Short Answer

Expert verified
The average value is \(\frac{1}{n+1}.\)

Step by step solution

01

Understand the Average Value Formula

The average value of a continuous function \(f(x)\) over the interval \([a, b]\) is given by \[\frac{1}{b-a} \int_{a}^{b} f(x) \, dx.\]Here, \(a = 0\) and \(b = 1\), so the formula simplifies to \[\int_{0}^{1} f(x) \, dx.\]
02

Set Up the Integral

We need to find the average value of \(f(x) = x^n\) over \([0, 1]\). Set up the integral:\[\frac{1}{1-0} \int_{0}^{1} x^n \, dx = \int_{0}^{1} x^n \, dx.\]
03

Integrate the Function

To evaluate the integral \(\int_{0}^{1} x^n \, dx\), use the power rule for integration, which states that \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C.\]Since we are evaluating a definite integral, we don't need the constant \(C\). Apply the limits:\[\left. \frac{x^{n+1}}{n+1} \right|_{0}^{1}.\]
04

Evaluate the Integral at the Limits

Substitute the limits into the evaluated integral:\[\frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1} - 0.\]
05

Conclusion

The average value of the function \(f(x) = x^n\) over the interval \([0, 1]\) is \(\frac{1}{n+1}.\)

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Key Concepts

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

Continuous Function
A continuous function is a fundamental concept in calculus and analysis. Imagine drawing a curve of the function on a graph without lifting your pen. This smoothness means there are no gaps, jumps, or sharp turns.
A continuous function's value changes very predictably and smoothly over its domain. This predictability is important because it allows us to perform operations like integration and differentiation with confidence.

For a function to be continuous at a point:
  • The function must be defined at that point.
  • The limit as it approaches the point from both directions must exist.
  • The function's value at the point must equal the limit of the function as it approaches the point.
In our exercise, the function \(f(x) = x^n\) is continuous over any real number interval, including \[0, 1\]. This assures us that we can calculate the average value using integration over this range without encountering mathematical interruptions.
Definite Integral
The definite integral is like the king of integration. It gives us the accumulated total of a function over a specific interval, providing a sort of 'sum' of what the function describes, considering both area and direction.
In mathematical terms, the definite integral is written as \[ \int_{a}^{b} f(x) \, dx \]. This notation means: calculate the integral of \(f(x)\) from \(x = a\) to \(x = b\).

During this exercise, integrating \(f(x) = x^n\) from \([0,1]\) computes the area under the curve of \(x^n\) between those two points on the x-axis, capturing the essence of the function over that range. The result helps us find the average value of the continuous function across the interval, which is also defined as \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]. This formula confirms that the contribution of each part of the interval to the total is equally considered, balancing out peaks and low points.
Power Rule for Integration
The power rule for integration is a handy shortcut for integrating polynomial functions. It simplifies the calculation of integrals, especially when dealing with powers of x.
The rule states: if you want to integrate \(x^n\) where eq -1\, the result is \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]. Here, \C\ is the constant of integration for indefinite integrals, representing the 'family' of antiderivatives.

For definite integrals, the constant \C\ isn't needed. After finding the antiderivative, we substitute the upper and lower limits of the interval directly into this expression. For our task, when solving \(\int_{0}^{1} x^n \, dx\), after applying the power rule, we replace \(x\) with 1 and 0 to get \[ \left. \frac{x^{n+1}}{n+1} \right|_{0}^{1} = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1} - 0 = \frac{1}{n+1} \].This process turns a potentially complex problem into a simple plug-and-play solution, demonstrating the beauty and utility of calculus in solving real-world and theoretical problems.

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Most popular questions from this chapter

An automobile dealer estimates that the newest model car will sell at the rate of \(30 / t\) cars per month, where \(t\) is measured in months and \(t=1\) corresponds to the beginning of January. Find the number of cars that will be sold from the beginning of January to the beginning of May.

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 \(2050(t=50)\).

BIOMEDICAL: Poiseuille's Law According to Poiseuille's law, the speed of blood in a blood vessel is given by \(V=\frac{p}{4 L v}\left(R^{2}-r^{2}\right)\) where \(R\) is the radius of the blood vessel, \(r\) is the distance of the blood from the center of the blood vessel, and \(p, L,\) and \(v\) are constants determined by the pressure and viscosity of the blood and the length of the vessel. The total blood flow is then given by $$ \left(\begin{array}{c} \text { Total } \\ \text { blood flow } \end{array}\right)=\int_{0}^{R} 2 \pi \frac{p}{4 L v}\left(R^{2}-r^{2}\right) r d r $$ Find the total blood flow by finding this integral \((p, L, v,\) and \(R\) are constants)

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). $$ \int(x-2)(x+4)^{5} d x $$

ECONOMICS: Pareto's Law The economist Vilfredo Pareto \((1848-1923)\) estimated that the number of people who have an income between \(A\) and \(B\) dollars \((A

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