Chapter 5: Problem 77
If a linear function passes through two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\), what is the average value of the function on the interval from \(x_{1}\) to \(x_{2}\) ?
Short Answer
Expert verified
The average value is \\(\frac{y_1+y_2}{2}\\) on interval \\[x_1, x_2\].
Step by step solution
01
Understand the Concept
The average value of a function on an interval \(a, b\) is calculated using the formula for the average value of a function: \frac{1}{b-a} \int_{a}^{b} f(x) \, dx\. For a linear function, this integral can be computed easily once we determine the function's equation.
02
Determine the Equation of the Linear Function
For a function passing through points \(x_{1}, y_{1}\) and \(x_{2}, y_{2}\), find the slope \(m\) using the formula \m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\. Then use the point-slope form \(y - y_{1} = m(x - x_{1})\) to express the linear function.
03
Integrate the Linear Function
The linear function can be expressed in the form \y = mx + c\. The integral from \(x_{1}\) to \(x_{2}\) would be \int_{x_{1}}^{x_{2}} (mx + c) \, dx\, which evaluates to \left[\frac{m}{2}x^{2} + cx\right]_{x_{1}}^{x_{2}}\.
04
Calculate the Definite Integral
Substitute the limits \(x_{1}\) and \(x_{2}\) into \frac{m}{2}x^{2} + cx\: \[\left(\frac{m}{2}x_{2}^{2} + cx_{2}\right) - \left(\frac{m}{2}x_{1}^{2} + cx_{1}\right)\] to find the definite integral.
05
Compute the Average Value
Divide the result of the integral by the length of the interval \(x_{2} - x_{1}\). This gives the average value of the linear function over the interval \(x_{1}, x_{2}\): \frac{1}{x_{2} - x_{1}}\left[(\frac{m}{2}x_{2}^{2} + cx_{2}) - (\frac{m}{2}x_{1}^{2} + cx_{1})\right]\.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Function
A linear function is a type of function where the relationship between the input (commonly denoted as \(x\)) and the output (commonly denoted as \(y\)) is a straight line when graphed on the Cartesian plane. The equation of a linear function can be represented in the general form \(y = mx + c\), where:
To find the equation of a linear function that passes through two given points \((x_1, y_1)\) and \((x_2, y_2)\), you first determine the slope \(m\) using the formula \(m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\). Once the slope is known, you can use the point-slope form to establish the complete equation of the line.
- \(m\) is the slope of the line, indicating how steep the line is. A positive slope means the line rises as \(x\) increases, and a negative slope means it falls.
- \(c\) is the y-intercept, the point where the line crosses the y-axis.
To find the equation of a linear function that passes through two given points \((x_1, y_1)\) and \((x_2, y_2)\), you first determine the slope \(m\) using the formula \(m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\). Once the slope is known, you can use the point-slope form to establish the complete equation of the line.
Definite Integral
A definite integral is a fundamental concept in calculus used to calculate the net area under a curve on a given interval. For a function \(f(x)\) defined over an interval \([a, b]\), the definite integral is represented as:\[\int_{a}^{b} f(x) \, dx\]This notation essentially sums up all the infinitely small values of \(f(x)\) from \(a\) to \(b\), resulting in the total area under the curve between these points.
- The process typically involves finding the antiderivative of \(f(x)\), which reverses differentiation.
- You then evaluate this antiderivative at the upper and lower bounds \(b\) and \(a\), subtracting the two results to get the net area.
Point-Slope Form
The point-slope form of a line is particularly useful when you know one point on the line and the slope. It provides a way to write the equation of a line straightforwardly. The point-slope form is given by:\[y - y_{1} = m(x - x_{1})\]Here,
Once you express the equation in point-slope form, you can reorganize it into the slope-intercept form \(y = mx + c\) if needed. Understanding and using point-slope form is critical when working with linear equations, especially in problems requiring the derivation of a line's equation given a slope and point.
- \((x_{1}, y_{1})\) represents a known point on the line.
- \(m\) is the slope of the line, as calculated between two points.
Once you express the equation in point-slope form, you can reorganize it into the slope-intercept form \(y = mx + c\) if needed. Understanding and using point-slope form is critical when working with linear equations, especially in problems requiring the derivation of a line's equation given a slope and point.