Problem 14
Question
Find an approximation of the area of the region \(R\) under the graph of the function \(f\) on the interval \([a, b] .\) In each case, use \(n\) subintervals and choose the representative points as indicated. \(f(x)=4-x^{2} ;[-1,2] ; n=6 ;\) left endpoints
Step-by-Step Solution
Verified Answer
The approximated area of the region R under the graph of the function \(f(x) = 4 - x^2\) on the interval [-1, 2] using 6 subintervals and left endpoints is \(A = \frac{17}{2}\).
1Step 1: Calculate the width of each subinterval
Calculate the interval length [a,b]:
\( \Delta x = \frac{(b-a)}{n} = \frac{(2-(-1))}{6} = \frac{3}{6} = \frac{1}{2}\)
2Step 2: Find the left endpoints of each subinterval
Determine the left endpoints for each of the 6 subintervals:
\(-1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}\)
3Step 3: Evaluate the function at each left endpoint
Evaluate the function \(f(x) = 4-x^2\) at the left endpoints:
\(f(-1) = 4-(-1)^2 = 4-1 = 3\)
\(f(-\frac{1}{2}) = 4-\left(-\frac{1}{2}\right)^2 = 4-\frac{1}{4} = \frac{15}{4}\)
\(f(0) = 4-0^2 = 4\)
\(f(\frac{1}{2}) = 4-\left(\frac{1}{2}\right)^2 = 4-\frac{1}{4} = \frac{15}{4}\)
\(f(1) = 4-1^2 = 4-1 = 3\)
\(f(\frac{3}{2}) = 4-\left(\frac{3}{2}\right)^2 = 4-\frac{9}{4} =\frac{7}{4}\)
4Step 4: Multiply each function value by the width of each subinterval to get the approximated area
Multiply the function value by the width of each subinterval (\(\Delta x = \frac{1}{2}\)):
\[A_1 = 3 \cdot \frac{1}{2} = \frac{3}{2}\]
\[A_2 = \frac{15}{4} \cdot \frac{1}{2} = \frac{15}{8}\]
\[A_3 = 4 \cdot \frac{1}{2} = 2\]
\[A_4 = \frac{15}{4} \cdot \frac{1}{2} = \frac{15}{8}\]
\[A_5 = 3 \cdot \frac{1}{2} = \frac{3}{2}\]
\[A_6 = \frac{7}{4} \cdot \frac{1}{2} = \frac{7}{8}\]
5Step 5: Add the areas to get the total approximated area
Add the areas to find the total approximated area under the curve:
\[A = A_1 + A_2 + A_3 + A_4 + A_5 + A_6\]
\[A = \frac{3}{2} + \frac{15}{8} + 2 + \frac{15}{8} +\frac{3}{2} + \frac{7}{8}\]
\[A = \frac{3}{2} + \frac{15}{8} + \frac{16}{8} + \frac{15}{8} +\frac{12}{8} + \frac{7}{8}\]
\[A = \frac{68}{8}\]
\[A = \frac{17}{2}\]
The approximated area of the region R under the graph of the function \(f(x) = 4 - x^2\) on the interval [-1, 2] using 6 subintervals and left endpoints is \(A = \frac{17}{2}\).
Key Concepts
Understanding ApproximationThe Role of SubintervalsUsing Left Endpoints in Riemann Sums
Understanding Approximation
Approximation is a method used to estimate a value that is difficult to measure directly. In the context of integrating functions, approximation helps find the area under a curve when an exact solution may not be possible or practical.
The Riemann sum is commonly used for this purpose. Here, the area under the curve is divided into smaller shapes (like rectangles), whose areas are easier to calculate.
Calculating each of these smaller areas gives an approximate total area under the function. In our example, we used rectangles based on the function \(f(x) = 4-x^2\) over the interval \([-1,2]\).
Although it may not be exactly correct, this method provides a close estimate, especially as the number of subintervals increases.
The Riemann sum is commonly used for this purpose. Here, the area under the curve is divided into smaller shapes (like rectangles), whose areas are easier to calculate.
Calculating each of these smaller areas gives an approximate total area under the function. In our example, we used rectangles based on the function \(f(x) = 4-x^2\) over the interval \([-1,2]\).
Although it may not be exactly correct, this method provides a close estimate, especially as the number of subintervals increases.
The Role of Subintervals
Subintervals are smaller segments into which the main interval is divided. In a Riemann sum, the number of subintervals \(n\) determines how many rectangles or shapes we'll use to approximate the area.
Subintervals allow for a more precise approximation, as each section can more accurately hug the curve of the function. We divided the interval \([-1,2]\) into 6 subintervals, giving us pieces of width \(\Delta x = \frac{1}{2}\).
More subintervals mean smaller \(\Delta x\), leading to a better approximation of the area since the smaller rectangles fit the curve more closely. This is why increasing \(n\) generally improves the accuracy.
Subintervals allow for a more precise approximation, as each section can more accurately hug the curve of the function. We divided the interval \([-1,2]\) into 6 subintervals, giving us pieces of width \(\Delta x = \frac{1}{2}\).
More subintervals mean smaller \(\Delta x\), leading to a better approximation of the area since the smaller rectangles fit the curve more closely. This is why increasing \(n\) generally improves the accuracy.
Using Left Endpoints in Riemann Sums
When calculating the Riemann sum, selecting representative points within each subinterval is key. One common method is using left endpoints, where we calculate the function's value at the leftmost point of each subinterval.
Here, the left endpoints were \(-1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}\). We used these values in our function \(f(x)\) to find the height of each rectangle.
Using left endpoints means our rectangles might sometimes undershoot or overshoot the true curve, especially at the beginning or end of intervals. However, this approach is systematic and straightforward, making it a popular choice for approximation.
As we saw, multiplying each function value by \(\Delta x\) and summing them provided an estimated area under the curve.
Here, the left endpoints were \(-1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}\). We used these values in our function \(f(x)\) to find the height of each rectangle.
Using left endpoints means our rectangles might sometimes undershoot or overshoot the true curve, especially at the beginning or end of intervals. However, this approach is systematic and straightforward, making it a popular choice for approximation.
As we saw, multiplying each function value by \(\Delta x\) and summing them provided an estimated area under the curve.
Other exercises in this chapter
Problem 14
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