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Chapter 5: Problem 3

Find the average value of the function over the given interval. $$f(x)=3 x ;[1,3]$$

Short Answer

Expert verified
The average value of the function over the interval is 6.

Step by step solution

01

Identify the Interval and Function

We are given the function \(f(x) = 3x\) and the interval \([1, 3]\). The first step is to understand that the average value of a function over a specific interval can be calculated using the formula for the average value of a function on an interval \([a, b]\), which is \(\frac{1}{b-a} \int_a^b f(x) \, dx\).
02

Set Up the Integral for the Average Value

Using the formula for the average value, we set up the integral expression:\[\text{Average} = \frac{1}{3-1} \int_1^3 3x \, dx\]Which simplifies to:\[\frac{1}{2} \int_1^3 3x \, dx.\]
03

Evaluate the Integral

Now we calculate the integral:\[\int_1^3 3x \, dx.\]To evaluate this, use the power rule for integration:\[\int 3x \, dx = \frac{3x^2}{2} + C.\]Apply the limits 1 and 3:\[\left[ \frac{3(3)^2}{2} - \frac{3(1)^2}{2} \right] = \frac{27}{2} - \frac{3}{2} = \frac{24}{2} = 12.\]
04

Find the Average Value

Now that we have the integral result, apply it to the average value formula:\[\text{Average} = \frac{1}{2} \times 12 = 6.\]
05

Conclusion

The average value of the function \(f(x) = 3x\) over the interval \([1, 3]\) is 6.

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Key Concepts

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

Integration
Integration is one of the fundamental concepts in calculus, used to find the area under a curve, among other applications. It is considered the reverse process of differentiation. While differentiation involves breaking things down into infinitesimally small pieces, integration adds these pieces together.
  • Integration can be thought of as summing up an infinite number of small quantities, usually represented as areas under a curve.
  • The integral sign, denoted by \(\int\), is much like an elongated 's' for 'sum'.
There are two main types of integrals: indefinite and definite integrals. The indefinite integral represents a family of functions characterized by a constant \(C\), whereas a definite integral computes a specific numerical value representing an area.
Power Rule for Integration
The power rule for integration is a straightforward method used to integrate functions that have the form \(x^n\). This rule states that if you have a function \(x^n\), its integral would be \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

Here's a simple breakdown:
  • Increase the power of \(x\) by 1.
  • Divide the term by this new power.
  • Don't forget to add the constant \(C\), which represents any constant value that could have been differentiated to give zero.
In our given problem, we employed the power rule for the function \(3x\). So, by applying \(\int 3x \, dx = \frac{3x^2}{2} + C\), we successfully integrated the function.
Definite Integral
The definite integral is crucial for finding the exact area under a curve over a specified interval. Unlike indefinite integrals, definite integrals yield numerical values. Here's how it works:
  • The notation for a definite integral includes limits of integration, typically shown as numbers at the top and bottom of the integral sign.
  • To evaluate, you first find the indefinite integral, then apply the upper and lower limits to find the difference, \(F(b) - F(a)\).
In the exercise we're discussing, after applying power rule, we plugged in the limits 1 and 3 into \(\frac{3x^2}{2}\) to find the total value for the definite integral, which resulted in 12 when computed.
Interval Analysis
Interval analysis involves focusing on a specific range for which we calculate or analyze a function. This is particularly important in calculus as it helps in determining limits and areas for real-world applications.
  • In our problem, the interval given is \([1, 3]\). This tells us we only consider the behavior and values of our function within this span.
  • Interval analysis sets the boundary conditions for our calculations, essentially sandwiching our area of interest.
By specifying the interval, we effectively limited our integral evaluation, ensuring the average value specifically pertains to \([1, 3]\). The length of the interval, denoted as \(b-a\), forms the denominator when using the formula for average value of a function. In this particular problem, it was \(3-1=2\), guiding the division to find the average.

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Most popular questions from this chapter

(a) The accompanying table shows the fraction of the Moon that is illuminated (as seen from Earth) at midnight (Eastern Standard Time) for the first week of 2005. Find the average fraction of the Moon illuminated during the first week of \(2005 .\) (b) The function \(f(x)=0.5+0.5 \sin (0.213 x+2.481)\) models data for illumination of the Moon for the first 60 days of \(2005 .\) Find the average value of this illumination function over the interval [0,7]. $$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text { DAY } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\\hline \text { ILLUMINATION } & 0.74 & 0.65 & 0.56 & 0.45 & 0.35 & 0.25 & 0.16 \\\\\hline\end{array}$$

Evaluate the limit by expressing it as a definite integral over the interval \([a, b]\) and applying appropriate formulas from geometry. $$\lim _{{\max }{\Delta x_{k} \rightarrow 0}}{\sum_{k=1}^{n} \sqrt{4-\left(x_{k}^{*}\right)^{2}}} \Delta x_{k} ; a=-2, b=2$$

(a) Evaluate the integral \(\int \sin x \cos x \, d x\) by two methods: first by letting \(u=\sin x,\) and then by letting \(u=\cos x\) (b) Explain why the two apparently different answers obtained in part (a) are really equivalent.

Find Find $$ \frac{d}{d x}\left[\int_{3 x}^{x^{2}} \frac{x-1}{1^{2}+1} d x\right] $$ by writing $$ \int_{3 x}^{x^{2}} \frac{t-1}{t^{2}+1} d t=\int_{3 x}^{0} \frac{t-1}{t^{2}+1} d t+\int_{0}^{x^{2}} \frac{t-1}{t^{2}+1} d t $$

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=1}^{100}\left(2^{k+1}-2^{k}\right)$$
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