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Magazine Chapter 8: Problem 4
Draw the graph of the function \(f(x)=2 x-1\) for \(-3
Short Answer
Expert verified
The integral evaluates to -6.
Step by step solution
01
Identifying the Function
The function given is a linear function of the form \(f(x) = 2x - 1\). The graph of this function is a straight line with a slope of 2 and a y-intercept at -1.
02
Plotting the Graph
To graph the linear function \(f(x) = 2x - 1\), first find the y-values for several x-values within the range \(-3 < x < 3\). For example, when \(x = -3\), \(f(x) = 2(-3) - 1 = -7\). When \(x = 0\), \(f(x) = -1\). And when \(x = 3\), \(f(x) = 5\). Plot these points and draw a straight line through them to represent the function.
03
Understanding the Integral
The integral \(\int_{-3}^{3}(2x - 1)\,dx\) represents the area under the curve of \(f(x) = 2x - 1\) from \(x = -3\) to \(x = 3\). This area can be positive, negative, or zero, depending on the function's position relative to the x-axis.
04
Calculating the Integral
Integrate the function \(f(x) = 2x - 1\) with respect to \(x\):\[\int (2x - 1)\,dx = x^2 - x + C\]Evaluate this indefinite integral from \(-3\) to \(3\):\[\left[ x^2 - x \right]_{-3}^{3} = (3^2 - 3) - ((-3)^2 - (-3))\]Calculate:\[ (9 - 3) - (9 + 3) = 6 - 12 = -6\]
05
Conclusion
The integral \(\int_{-3}^{3}(2x - 1)\,dx\) evaluates to -6, meaning the net area between the curve and the x-axis over this interval is -6. This indicates more area is below the x-axis than above within this range.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Functions
Linear functions are a fundamental concept in mathematics. A linear function is an algebraic equation in which the highest degree of the variable is one. The general form is:
\( f(x) = mx + b \)
Here:
When dealing with linear functions, we are describing a relationship where each increase or decrease in \(x\) results in a consistent change in \(f(x)\). Understanding this helps in swiftly sketching them, as we shall see in the next section.
\( f(x) = mx + b \)
Here:
- \(m\) is the slope of the line.
- \(b\) is the y-intercept, where the line crosses the y-axis.
When dealing with linear functions, we are describing a relationship where each increase or decrease in \(x\) results in a consistent change in \(f(x)\). Understanding this helps in swiftly sketching them, as we shall see in the next section.
Graph Sketching
Graph sketching for a linear function is relatively straightforward. The function given, \(f(x) = 2x - 1\), provides all necessary information:
Employing these steps makes it easy to draw the graph of any linear function efficiently. Observing the changes in y as x shifts helps in verifying the accuracy of your graph.
- The slope \(m = 2\) tells us the line rises two units for each unit move to the right.
- The y-intercept is -1, meaning the line crosses the y-axis at point (0, -1).
Employing these steps makes it easy to draw the graph of any linear function efficiently. Observing the changes in y as x shifts helps in verifying the accuracy of your graph.
Definite Integrals
Definite integrals in calculus denote the signed area under a curve within an interval on the x-axis. For the function \(f(x) = 2x - 1\), we're interested in the area between \(-3\) and \(3\).
The integral \(\int_{-3}^{3}(2x - 1)\,dx\) calculates this area:
The result of -6 indicates that the net area below the x-axis is greater within this range, emphasizing how definite integrals calculate net surface area, factoring in above and below the axis contributions.
The integral \(\int_{-3}^{3}(2x - 1)\,dx\) calculates this area:
- First integrate to find the antiderivative: \(x^2 - x + C\).
- Evaluate this antiderivative from \(-3\) to \(3\):
- Substitute 3 into the antiderivative: \((3^2 - 3) = 6\).
- Substitute -3: \(((-3)^2 + 3) = 12\).
- Subtract the results: \(6 - 12 = -6\).
The result of -6 indicates that the net area below the x-axis is greater within this range, emphasizing how definite integrals calculate net surface area, factoring in above and below the axis contributions.
