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Magazine Chapter 5: Problem 5
A graph of a function \(f(x)\) is given. Using the geometry of the graph, evaluate the definite integrals. (a) \(\int_{0}^{1}(-2 x+4) d x\) (d) \(\int_{1}^{3}(-2 x+4) d x\) (b) \(\int_{0}^{2}(-2 x+4) d x\) (e) \(\int_{2}^{4}(-2 x+4) d x\) (c) \(\int_{0}^{3}(-2 x+4) d x\) (f) \(\int_{0}^{1}(-6 x+12) d x\)
Short Answer
Expert verified
The integrals are: (a) 3, (b) 4, (c) 3, (d) 0, (e) -4, (f) 6.
Step by step solution
01
Identify the function and graph type
The given function is a linear function \( f(x) = -2x + 4 \). The graph of this function is a straight line. The slope is \(-2\), which means the line is decreasing, and it intersects the y-axis at \(4\). We will evaluate the area (integral) under this line between different intervals.
02
Calculate area for \(a\) The integral from 0 to 1
We need to find the area under the line \(f(x) = -2x + 4\) from \(x=0\) to \(x=1\). The line crosses the y-axis at \(y = 4\) and the x-axis at \(x = 2\). This forms a right triangle between \(x=0\) and \(x=1\) with height \(4\) at \(x=0\) and height \(2\) at \(x=1\).\[\text{Area} = \text{(Base) \(\cdot\) (Height)}/2 = 1 \cdot (4+2)/2 = 3\] So, the value of \(\int_{0}^{1}(-2 x+4) dx\) is \(3\).
03
Calculate area for \(b\) The integral from 0 to 2
For \(x=0\) to \(x=2\), our triangle will stretch from the y-axis intercept to the x-axis intercept. The height at \(x=0\) is 4, and reaching 0 at \(x=2\).\[\text{Area} = (\text{Base} \cdot \text{Height})/2 = 2 \cdot 4/2 = 4\] So, \(\int_{0}^{2}(-2 x+4) dx = 4\).
04
Calculate area for \(c\) The integral from 0 to 3
From \(x=0\) to \(x=3\), the area is first a triangle from \(x=0\) to \(x=2\) as calculated earlier, which is \(4\), and then a triangle from \(x=2\) to \(x=3\) with the base length \(1\), height from line at \(x=2\) (which is 0) to the height below x-axis (down to -2 at \(x=3\)).\[\text{Additional negative area} = (\text{Base} \cdot \text{Height}) / 2 = 1 \cdot 2 / 2 = -1\]Total area is \(4 + (-1) = 3\).So, \(\int_{0}^{3}(-2 x+4) dx = 3\).
05
Calculate area for \(d\) The integral from 1 to 3
From \(x=1\) to \(x=3\), it compares a trapezoid and a triangle: at \(x=1\), the height is \(2\), reaching 0 at \(x=2\) and -2 at \(x=3\). Thus its forms a triangle \(x=2\) to \(x=3\) as previously noted, and a right trapezoid from \(x=1\) to \(x=3\).\[\text{Area of trapezoid} = ((2 + 0) / 2 \cdot 1) = 1\]\[\text{Total area} = 1 - 1 = 0\]\(\int_{1}^{3}(-2 x+4) dx = 0\).
06
Calculate area for \(e\) The integral from 2 to 4
From \(x=2\) to \(x=3\) is a previous triangle with an area \(-1\), then from \(x=3\) to \(x=4\) the height goes from \(-2\) to \(-4\).\[\text{Area of trapezoid} = ((-2 + -4)/2 \cdot 1) = -3\]Total of negative areas is \(-1 + -3 = -4\).So, \(\int_{2}^{4}(-2 x+4) dx = -4\).
07
Calculate area for \(f\) The integral from 0 to 1 of a new function
This time for \(f(x) = -6x + 12\) from \(x=0\) to \(x=1\), the y-intercept at \(12\), line goes through \(x=2\) as x-intercept. Forms a triangle with height \(6\) from \(x=0\) to \(x=1\).\[\text{Area} = (1 \cdot (12 - 6)/2) = 6\]So, \(\int_{0}^{1}(-6 x+12) dx = 6\).
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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 among the simplest types of functions and are characterized by their straight-line graphs on the coordinate plane. The general form of a linear function is \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept. In the function \( f(x) = -2x + 4 \), the slope is \(-2\) and the y-intercept is \(4\). This means the graph of this function is a straight line that decreases as you move from left to right.
- The slope \(-2\) indicates that for every one unit you move right on the x-axis, the line moves two units down on the y-axis.
- The y-intercept \(4\) is the point where the line crosses the y-axis. This tells us the value of \( y \) when \( x \) is zero.
Area Under a Curve
The area under a curve in a graph is a crucial concept in calculus, especially when dealing with integrals. For linear functions like \( f(x) = -2x + 4 \), this area often forms simple geometric shapes such as triangles or trapezoids. Calculating the area under a curve involves integrating the function over the desired interval.
- For the integral \( \int_{0}^{1}(-2 x+4) dx \), the area under the curve from \( x=0 \) to \( x=1 \) forms a triangle with a calculable area.
- From the intersection of the line and the x-axis, or when the function dips below the x-axis, the calculated area might include negative areas, indicating regions below this axis.
Integral Calculus
Integral Calculus is a powerful mathematical tool that allows us to find accumulation functions, such as areas under curves. In definite integrals, we look to find the total area between the curve and the x-axis over a specified interval. This is represented by the notation:\[ \int_{a}^{b} f(x) \ dx \]Where \(a\) and \(b\) are the limits of integration defining the section of the curve we wish to analyze.
- The value obtained from evaluating a definite integral represents the signed area under the curve between these limits.
- If the curve lies below the x-axis between any part of the limits, that portion contributes negatively to the total integral value.
Geometry of Graphs
The geometry of graphs involves understanding the spatial representation of functions and their relationships to the coordinate plane. When dealing with linear functions like \( f(x) = -2x + 4 \), the graph is a line whose properties are defined by the equation. The intersection points, intercepts, and slopes are all key geometric features.
- Linear functions have geometric properties that allow for easy calculation of areas under the curve using simple shapes like triangles and trapezoids.
- The intersection between the line and the x-axis can help identify points where the function shape changes, often splitting the graph into sections of positive and negative areas.
