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Magazine Chapter 5: Problem 15
\(\approx\) In Problems \(11-28\), sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. $$ y=\frac{1}{4}\left(x^{2}-7\right), y=0, \text { between } x=0 \text { and } x=2 $$
Short Answer
Expert verified
The area of the region is \(\frac{17}{6}\) square units.
Step by step solution
01
Sketch the Region
The equations given are \(y = \frac{1}{4}(x^2 - 7)\) and \(y = 0\). The region of interest is between \(x = 0\) and \(x = 2\). Plot both equations on the Cartesian plane. The parabola \(y = \frac{1}{4}(x^2 - 7)\) is a downward-opening parabola shifted down by 7. The line \(y = 0\) represents the x-axis. Identify the region that lies between these curves from \(x = 0\) to \(x = 2\).
02
Identify a Typical Slice
To approximate the area of the given region, consider a vertical slice at position \(x\), which is a small rectangle with width \(\Delta x\) and height equal to the difference between the top and bottom functions: \(h = \frac{1}{4}(x^2 - 7) - 0 = \frac{1}{4}(x^2 - 7)\).
03
Approximate the Area of Slices
The area of one such slice is the height \(\frac{1}{4}(x^2 - 7)\) times the width \(\Delta x\). Summing up the contributions from all such slices between \(x = 0\) and \(x = 2\) approximates the area. This leads to the Riemann sum: \(\sum \frac{1}{4}(x_i^2 - 7)\Delta x\).
04
Set Up the Integral
To compute the exact area, set up the definite integral based on the Riemann sum approximation. The area is given by the integral: \[A = \int_0^2 \frac{1}{4}(x^2 - 7) \, dx\]
05
Simplify and Integrate
First, simplify the integral: \[A = \frac{1}{4} \int_0^2 (x^2 - 7) \, dx = \frac{1}{4} \left( \int_0^2 x^2 \, dx - \int_0^2 7 \, dx \right)\]Calculate each integral separately: For \(\int_0^2 x^2 \, dx\), use the antiderivative \(\frac{1}{3}x^3\) to get:\[\left[\frac{1}{3}x^3\right]_0^2 = \frac{1}{3}(2^3) - \frac{1}{3}(0^3) = \frac{8}{3}\]For \(\int_0^2 7 \, dx\), use the antiderivative \(7x\) to get:\[\left[7x\right]_0^2 = 7(2) - 7(0) = 14\]Substitute back:\[A = \frac{1}{4} \left( \frac{8}{3} - 14 \right) = \frac{1}{4} \left( \frac{8 - 42}{3} \right) = \frac{1}{4} \times \left( -\frac{34}{3} \right) = -\frac{34}{12} = -\frac{17}{6} \]
06
Confirm and Interpret the Results
Since area cannot be negative, interpret the absolute value for context. Thus, the final area is \(\frac{17}{6}\) square units. This is a good final verification step to ensure consistency with expectations of positive areas.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
In calculus, a definite integral is a powerful tool that helps us find the area under a curve and between two bounds on the x-axis. Imagine a plotted graph with an intriguing shape. Calculating the area under this shape is what definite integrals allow us to do.
Usually expressed as \( \int_a^b f(x) \, dx \), the definite integral finds the signed area bounded by the curve \(y = f(x)\), the x-axis, and vertical lines at \(x = a\) and \(x = b\). The bounds \(a\) and \(b\) are the 'limits of integration'. They tell us where our area calculation starts and ends on the x-axis.
A definite integral has several important applications:
Usually expressed as \( \int_a^b f(x) \, dx \), the definite integral finds the signed area bounded by the curve \(y = f(x)\), the x-axis, and vertical lines at \(x = a\) and \(x = b\). The bounds \(a\) and \(b\) are the 'limits of integration'. They tell us where our area calculation starts and ends on the x-axis.
A definite integral has several important applications:
- Calculating areas and volumes.
- Solving problems in physics related to work and energy.
- Finding displacement and accumulated quantities in various contexts.
Riemann Sum
The Riemann sum is a technique for estimating the area under a curve. Named after the mathematician Bernhard Riemann, this method approximates definite integrals, providing a bridge between discrete sums and continuous integration.
By dividing the area under a curve into a series of simple shapes, like rectangles, we can sum up their areas to estimate the total area. Specifically, imagine slicing the area between two points on a graph into very thin vertical slices, or rectangles. Each slice's height is determined by the function's value at a point within the slice, and its width is a small segment of the x-axis, denoted by \( \Delta x \).
The Riemann sum is then given by:
\[ \sum_{i=1}^{n} f(x_i^*) \Delta x \]
where \( f(x_i^*) \) is the function's value at a point \( x_i^* \) within the i-th interval. As \( \Delta x \) becomes smaller and the number of intervals \( n \) becomes larger, the Riemann sum approaches the exact value of the definite integral.
Key points about Riemann sums include:
By dividing the area under a curve into a series of simple shapes, like rectangles, we can sum up their areas to estimate the total area. Specifically, imagine slicing the area between two points on a graph into very thin vertical slices, or rectangles. Each slice's height is determined by the function's value at a point within the slice, and its width is a small segment of the x-axis, denoted by \( \Delta x \).
The Riemann sum is then given by:
\[ \sum_{i=1}^{n} f(x_i^*) \Delta x \]
where \( f(x_i^*) \) is the function's value at a point \( x_i^* \) within the i-th interval. As \( \Delta x \) becomes smaller and the number of intervals \( n \) becomes larger, the Riemann sum approaches the exact value of the definite integral.
Key points about Riemann sums include:
- They provide a way to transition from finite sums to continuous integrals.
- Useful for practical calculations when an antiderivative is difficult to find.
- Can be adapted to higher dimensions for volume calculations.
Area under Curves
Finding the area under curves is a fundamental application of integration. When you have a graph representing various phenomena, such as speed over time or pressure over volume, the area under the curve can give you valuable insights into the underlying behavior. For example, calculating the area under a velocity-time graph yields the total distance traveled.
The process of finding the area involves setting up a definite integral of the function that defines the curve. Consider the function \( y = f(x) \), where the area under the curve from \( x = a \) to \( x = b \) is given by the integral:
\[ \text{Area} = \int_a^b f(x) \, dx \]
Using this approach, you can find areas for simple linear functions or more complex curves like parabolas or sine waves.
Important aspects of finding areas using integrals:
The process of finding the area involves setting up a definite integral of the function that defines the curve. Consider the function \( y = f(x) \), where the area under the curve from \( x = a \) to \( x = b \) is given by the integral:
\[ \text{Area} = \int_a^b f(x) \, dx \]
Using this approach, you can find areas for simple linear functions or more complex curves like parabolas or sine waves.
Important aspects of finding areas using integrals:
- Definite integrals capture the total area between a curve and the x-axis within specified bounds.
- Areas can be positive or negative, based on whether the curve is above or below the x-axis.
- Integration allows for exact calculations where geometric methods fall short.
