Problem 5
Question
5\. (a) Estmate the area under the graph of \(f(x)=1+x^{2}\) from \(x=-1\) to \(x=2\) using three rectangles and right endpoints. Then improve your estimate by using six rectangles. Sketch the curve and the approximating rectangles. (b) Repeat part (a) using left endpoints. (b) Repeat part (a) using midpoints. (d) From your sketches in parts (a) - (c), which appears to be the best estimate?
Step-by-Step Solution
Verified Answer
The midpoint method with 6 rectangles provides the best estimate.
1Step 1: Understanding the Problem
We need to estimate the area under the curve of the function \(f(x) = 1 + x^2\) from \(x = -1\) to \(x = 2\) using rectangles. We will use three different methods: right endpoints, left endpoints, and midpoints, each with three and six rectangles. Finally, we'll determine which method provides the best estimation.
2Step 1: Using Right Endpoints with 3 Rectangles
Divide the interval \([-1, 2]\) into 3 equal parts: \([-1, 0)\), \([0, 1)\), and \([1, 2]\). Each part has a width \(\Delta x = 1\). Calculate the height for each rectangle using the right endpoint of each interval: - Interval \([-1, 0)\): right endpoint \(x_1 = 0\), height \(f(0) = 1\).- Interval \([0, 1)\): right endpoint \(x_2 = 1\), height \(f(1) = 2\).- Interval \([1, 2]\): right endpoint \(x_3 = 2\), height \(f(2) = 5\).The area is the sum of areas of the rectangles: \(1\cdot(1) + 1\cdot(2) + 1\cdot(5) = 8\).
3Step 2: Using Right Endpoints with 6 Rectangles
Divide \([-1, 2]\) into 6 equal parts: each \(\Delta x = 0.5\), yielding subintervals \([-1, -0.5)\), \([-0.5, 0)\), \([0, 0.5)\), \([0.5, 1)\), \([1, 1.5)\), \([1.5, 2]\). Heights are determined using right endpoints - Heights: \( f(-0.5), f(0), f(0.5), f(1), f(1.5), f(2) \)- Calculations: \( f(-0.5) = 1.25 \), \( f(0) = 1 \), \( f(0.5) = 1.25 \), \( f(1) = 2 \), \( f(1.5) = 3.25 \), \( f(2) = 5 \).Area = \(0.5(1.25 + 1 + 1.25 + 2 + 3.25 + 5) = 6.875\).
4Step 3: Using Left Endpoints with 3 Rectangles
Divide \([-1, 2]\) into 3 parts, using left endpoints: \([-1, 0)\), \([0, 1)\), \([1, 2)\). Calculating heights:- Heights: \(f(-1), f(0), f(1)\)- Calculate: \( f(-1)=2, f(0)=1, f(1)=2 \).Area = \(1(2) + 1(1) + 1(2) = 5\).
5Step 4: Using Left Endpoints with 6 Rectangles
Divide \([-1, 2]\) into 6 parts, use left endpoints: \([-1, -0.5)\), \([-0.5, 0)\), \([0, 0.5)\), \([0.5, 1)\), \([1, 1.5)\), \([1.5, 2)\). Calculate heights:- Heights: \( f(-1), f(-0.5), f(0), f(0.5), f(1), f(1.5) \)- Calculate: \( f(-1)=2, f(-0.5)=1.25, f(0)=1, f(0.5)=1.25, f(1)=2, f(1.5)=3.25 \).Area = \(0.5(2 + 1.25 + 1 + 1.25 + 2 + 3.25) = 5.875\).
6Step 5: Using Midpoints with 3 Rectangles
Subdivide into 3 intervals and use midpoints: - Midpoints for \([-1, 0)\), \([0, 1)\), \([1, 2)\) are \(-0.5, 0.5, 1.5\).- Heights: - \( f(-0.5)=1.25 \), - \( f(0.5)=1.25 \), - \( f(1.5)=3.25 \).Area = \(1.25 + 1.25 + 3.25 = 5.75\).
7Step 6: Using Midpoints with 6 Rectangles
Subdivide into 6 intervals and use midpoints - Midpoints: \(-0.75, -0.25, 0.25, 0.75, 1.25, 1.75\)- Heights: - \( f(-0.75)=1.5625 \), - \( f(-0.25)=1.0625 \), - \( f(0.25)=1.0625 \), - \( f(0.75)=1.5625 \), - \( f(1.25)=2.5625 \), - \( f(1.75)=4.0625 \).Area = \(0.5(1.5625 + 1.0625 + 1.0625 + 1.5625 + 2.5625 + 4.0625) = 6.46875\).
8Step 7: Compare Estimates and Conclude
Comparing all estimations for area:- Right endpoints give \(8\) with 3 and \(6.875\) with 6 rectangles- Left endpoints give \(5\) with 3 and \(5.875\) with 6 rectangles- Midpoints give \(5.75\) with 3 and \(6.46875\) with 6 rectanglesMidpoint method with 6 rectangles appears to provide the best approximation.
Key Concepts
Definite IntegralsApproximation MethodsFunction Graphing
Definite Integrals
Definite integrals are a fundamental concept in calculus. They provide a way to calculate the area under the curve of a function over a specific interval. For students, understanding definite integrals is crucial as they extend the idea of summation and limits, which form the basis of calculus.
To compute a definite integral, you integrate a function between two limits: the lower and upper bounds of the interval. For example, in the problem you've been solving, the function was \(f(x) = 1 + x^2\), with limits from \(x = -1\) to \(x = 2\). The process involves finding the difference \(F(b) - F(a)\), where \(F\) is any antiderivative of \(f\).
When you use rectangles to approximate these areas, like in the exercise, it's a practical way to visualize what a definite integral calculates. It involves slicing the total area into small pieces, combining all these slivers to approximate the total area under the curve. The more slices or rectangles used, the closer you approximate the true area, aligning with how definite integrals work.
To compute a definite integral, you integrate a function between two limits: the lower and upper bounds of the interval. For example, in the problem you've been solving, the function was \(f(x) = 1 + x^2\), with limits from \(x = -1\) to \(x = 2\). The process involves finding the difference \(F(b) - F(a)\), where \(F\) is any antiderivative of \(f\).
When you use rectangles to approximate these areas, like in the exercise, it's a practical way to visualize what a definite integral calculates. It involves slicing the total area into small pieces, combining all these slivers to approximate the total area under the curve. The more slices or rectangles used, the closer you approximate the true area, aligning with how definite integrals work.
Approximation Methods
There are several methods to approximate the area under a curve, each providing a different estimation. In the context of the problem above, three methods are used: right endpoint, left endpoint, and midpoint rules. These are popular techniques associated with Riemann Sums, named after the mathematician Bernhard Riemann.
1. **Right Endpoint Method:** Each rectangle's height is determined by the function's value at the right endpoint of each subinterval. This method can sometimes overestimate or underestimate the area, depending on the function's slope in the interval.
2. **Left Endpoint Method:** The function's value at the left endpoint of each subinterval sets the height of each rectangle. Like the right endpoint method, this can lead to an over or underestimation, depending on the curve's direction.
3. **Midpoint Method:** This approach uses the value of the function at the midpoint of each subinterval. It typically provides a more balanced approximation since it considers values near the middle of the interval, often leading to more accurate results than using just the endpoints.
These approximation methods highlight how calculus tools can be applied in solving real-world problems, giving students a foundation for more complex calculus concepts.
1. **Right Endpoint Method:** Each rectangle's height is determined by the function's value at the right endpoint of each subinterval. This method can sometimes overestimate or underestimate the area, depending on the function's slope in the interval.
2. **Left Endpoint Method:** The function's value at the left endpoint of each subinterval sets the height of each rectangle. Like the right endpoint method, this can lead to an over or underestimation, depending on the curve's direction.
3. **Midpoint Method:** This approach uses the value of the function at the midpoint of each subinterval. It typically provides a more balanced approximation since it considers values near the middle of the interval, often leading to more accurate results than using just the endpoints.
These approximation methods highlight how calculus tools can be applied in solving real-world problems, giving students a foundation for more complex calculus concepts.
Function Graphing
Graphing functions is a powerful visual tool in mathematics, particularly in calculus, to comprehend the behavior of a function over an interval. In this specific exercise, graphing \(f(x) = 1 + x^2\) and the approximating rectangles provides insight into how well each Riemann Sum method estimates the area under the curve.
Each function has unique characteristics that can be observed through graphing. For the function \(f(x) = 1 + x^2\), you have a parabola that opens upwards. This shape means any approximation method that relies purely on endpoints without adjusting for the function's curve can lead to significant discrepancies between actual and estimated values.
By sketching the rectangles, one can see how varying the size and number of rectangles influences the estimation's accuracy for the area under the graph. Smaller rectangles, meaning more rectangles over the same interval, generally result in a better fit and more precise approximation, closely following the function's path. Thus, graphing is not only a skill to be developed but also an essential step in verifying and understanding approximation methods.
Each function has unique characteristics that can be observed through graphing. For the function \(f(x) = 1 + x^2\), you have a parabola that opens upwards. This shape means any approximation method that relies purely on endpoints without adjusting for the function's curve can lead to significant discrepancies between actual and estimated values.
By sketching the rectangles, one can see how varying the size and number of rectangles influences the estimation's accuracy for the area under the graph. Smaller rectangles, meaning more rectangles over the same interval, generally result in a better fit and more precise approximation, closely following the function's path. Thus, graphing is not only a skill to be developed but also an essential step in verifying and understanding approximation methods.
Other exercises in this chapter
Problem 5
Evaluate the integral. \(\int_{0}^{1} x^{4 / 5} d x\)
View solution Problem 5
Evaluate the integral. \(\int r e^{r / 2} d r\)
View solution Problem 5
Evaluate the integral by making the given substitution. \(\int \cos ^{3} \theta \sin \theta d \theta, \quad u=\cos \theta\)
View solution Problem 6
Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \(\int_{0}^{\infty} \frac{x}{\left(x^{2}+2\right)^{2}} d x\)
View solution