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Show that the average value of a linear function \(L ( x )\) on \([ a , b ]\) is

Short Answer

Expert verified
The average value of a linear function \(L(x)\) on the interval \([a , b]\) is given by the arithmetic mean between its values at points \(a\) and \(b\), represented as \( AVG = \frac{1}{2}(L(a) + L(b)) \).

Step by step solution

01

Definition of the average value of a function

The main formula to utilize here is that the average value of a function \(f(x)\) on a closed interval \([a,b]\) is given as: \( AVG = {1 \over {b-a}} \int_a^b f(x) dx \)
02

Substitute the linear function

Given a linear function in the form \(L(x) = mx + n\), substitute into the formula: \( AVG = {1 \over {b-a}} \int_a^b (mx + n) dx \)
03

Compute the integral

On this step, compute the integral of the linear function. Usually, it is most efficient to break it down into two integrals for easier computation: \( AVG = {1 \over {b-a}} ( [ \frac{1}{2}m x^2 + nx]_a^b ) \)
04

Evaluate the integral

The integral, when evaluated from \(a\) to \(b\), results in: \( AVG = {1 \over {b-a}} ( \frac{1}{2}m b^2 + nb - \frac{1}{2}m a^2 - na ) \)
05

Simplify

Simplify the expression for the average value of the function. After the simplification and grouping of like terms, the expression ends with: \( AVG = \frac{1}{2}m(b+a) + n \)
06

Express the result using the values of L(a) and L(b)

The average value of the linear function is equivalent to the mean of the function's values at points \(a\) and \(b\). Therefore, \( AVG = \frac{1}{2}(L(a) + L(b)) \)

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Key Concepts

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

Linear Functions
A linear function is a type of function that creates a straight line when graphed on a coordinate plane. These functions have a constant slope and can be expressed in the standard form \( L(x) = mx + n \), where:\
    \
  • \( m \) is the slope, which measures the steepness of the line. It indicates how much \( y \) changes for a unit change in \( x \).
  • \
  • \( n \) is the y-intercept, representing the point where the line crosses the y-axis.
\Linear functions are simple yet powerful, as they depict relationships where changes in one variable consistently translate to proportional changes in another.
Understanding linear functions is crucial because they form the foundation for many higher-level concepts in calculus and other mathematical areas. Recognizing their graph, interpreting the slope and intercept, and using them in expressions and computations are useful skills in problem-solving.
Integral Calculus
Integral calculus is a branch of mathematics that deals with integration. Integration is a method used to compute areas, volumes, central points, and many other useful things.
It is the reverse process of differentiation, meaning while differentiation derives the instantaneous rate of change of a function, integration accumulates the quantities.
Notably, integration has two primary applications:
  • Definite Integrals: These calculate the total accumulation between two points, such as the total area under a curve between points \( a \) and \( b \).
  • Indefinite Integrals: These provide a function whose derivative is the original function, offering a general solution.
The act of integrating a function can be likened to summing an infinite number of infinitesimal changes, which provides a profound insight into the function's accumulated value over an interval.
Definite Integral
A definite integral is a concept in integral calculus that computes the accumulation of quantities, such as area, between a specific range. For a function \( f(x) \) over an interval \([a,b]\), the definite integral is given by:\[\int_a^b f(x) \ dx\]This expression represents the area under the curve of \( f(x) \) from \( x = a \) to \( x = b \). The result is a numerical value indicating total accumulation, which could be distance, area, or other measurable properties.
The process of finding a definite integral involves:
  • Identifying the function to be integrated.
  • Finding the antiderivative of the function.
  • Applying the limits \( a \) and \( b \).
  • Calculating the difference between the values at \( b \) and \( a \).
This is crucial in solving real-world problems where exact summation over an interval is necessary, such as calculating displacement over time or total income over a fiscal period. Understanding the definite integral's role bridges the abstract concept of integration with tangible results.

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Most popular questions from this chapter

In Exercises \(9-12,\) use RAM to estimate the area of the region enclosed between the graph of \(f\) and the \(x\) -axis for \(a \leq x \leq b\) . $$f(x)=\frac{1}{x}, \quad a=1, \quad b=3$$

In Exercises \(23-26\) use a calculator program to find the Simpson's Rule approximations with \(n=50\) and \(n=100 .\) $$\int_{0}^{1} \sqrt{1+x^{4}} d x$$

Multiple Choice An LRAM sum with 4 equal subdivisions is used to approximate the area under the sine curve from \(x=0\) to \(x=\pi .\) What is the approximation? (A) \(\frac{\pi}{4}\left(0+\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}\right)\) B) \(\frac{\pi}{4}\left(0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1\right)\) (C) \(\frac{\pi}{4}\left(0+\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2}\right)\) D) \(\frac{\pi}{4}\left(0+\frac{1}{2}+\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\right)\) (E) \(\frac{\pi}{4}\left(\frac{1}{2}+\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}+1\right)\)

Multiple Choice If \(f\) is a positive, continuous function on an interval \([a, b],\) which of the following rectangular approximation methods has a limit equal to the actual area under the curve from \(a\) to \(b\) as the number of rectangles approaches infinity? $$\begin{array}{c}{\text { I. LRAM }} \\ {\text { II. RRAM }} \\ {\text { III. MRAM }}\end{array}$$ (A) I and II only (B) III only (C) I and III only (D)I, II, and III (E) None of these

Multiple Choice If an MRAM sum with four rectangles of equal width is used to approximate the area enclosed between the \(x\) -axis and the graph of \(y=4 x-x^{2},\) the approximation is (A) 10 (B) 10.5 (C) \(10 . \overline{6}\) (D) 10.75 (E) 11

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