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Chapter 5: Problem 3

Use finite approximations to estimate the area under the graph of the function using a. a lower sum with two rectangles of equal width. b. a lower sum with four rectangles of equal width. c. an upper sum with two rectangles of equal width. d. an upper sum with four rectangles of equal width. \(f(x)=1 / x\) between \(x=1\) and \(x=5\)

Short Answer

Expert verified
Lower sums: \( \frac{8}{3}, \frac{25}{12} \); Upper sums: \( \frac{16}{15}, \frac{77}{60} \).

Step by step solution

01

Graph Analysis

We are tasked with estimating the area under the curve of the function \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = 5 \). We will use finite approximations by dividing this interval into either two or four rectangles for both lower and upper sum calculations.
02

Step A: Lower Sum with Two Rectangles

Divide the interval [1, 5] into two subintervals of equal width, each width being \( \Delta x = \frac{5-1}{2} = 2 \). For a lower sum, we will use the left endpoint of each subinterval to determine the height of the rectangle. Evaluate \( f(x) \) at \( x = 1 \) and \( x = 3 \):\[ f(1) = 1, \quad f(3) = \frac{1}{3} \]. The lower sum is \[ 2 \times f(1) + 2 \times f(3) = 2 \times 1 + 2 \times \frac{1}{3} = 2 + \frac{2}{3} = \frac{8}{3} \].
03

Step B: Lower Sum with Four Rectangles

Divide the interval [1, 5] into four smaller intervals each of width \( \Delta x = \frac{4}{4} = 1 \). Using left endpoints, evaluate \( f(x) \) at each \( x = 1, 2, 3, 4 \):\[ f(1) = 1, \quad f(2) = \frac{1}{2}, \quad f(3) = \frac{1}{3}, \quad f(4) = \frac{1}{4} \]. The lower sum is \[ 1 \times f(1) + 1 \times f(2) + 1 \times f(3) + 1 \times f(4) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{12} \].
04

Step C: Upper Sum with Two Rectangles

For the upper sum with two rectangles, use the right endpoints of each subinterval in [1, 5], which are \( x = 3 \) and \( x = 5 \). Evaluate \( f(x) \) at these points: \[ f(3) = \frac{1}{3}, \quad f(5) = \frac{1}{5} \]. Calculate the upper sum: \[ 2 \times f(3) + 2 \times f(5) = 2 \times \frac{1}{3} + 2 \times \frac{1}{5} = \frac{2}{3} + \frac{2}{5} = \frac{16}{15} \].
05

Step D: Upper Sum with Four Rectangles

Divide into four subintervals again, using right endpoints \( x = 2, 3, 4, 5 \). Evaluate \( f(x) \):\[ f(2) = \frac{1}{2}, \quad f(3) = \frac{1}{3}, \quad f(4) = \frac{1}{4}, \quad f(5) = \frac{1}{5} \]. The upper sum is \[ 1 \times f(2) + 1 \times f(3) + 1 \times f(4) + 1 \times f(5) = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{77}{60} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Upper and Lower Sums
When estimating areas under curves, using upper and lower sums provides a practical approach. This is particularly useful when seeking the area beneath a function like \( f(x) = \frac{1}{x} \) between two points, say \( x = 1 \) and \( x = 5 \).
Upper sums rely on the highest point of the function in a partition subinterval to determine the rectangle's height, thereby providing an overestimate of the area. Conversely, lower sums use the lowest point, leading to an underestimate.
This method involves calculating rectangles that exactly fit under (lower sum) or above (upper sum) the curve on divided sections of an interval. These calculations give us insights by providing bounding values on the actual area.
Finite Approximations
Finite approximations simplify complex integrations by breaking them into manageable sections. Consider the function \( f(x) = \frac{1}{x} \) over the interval \( [1, 5] \). By dividing this into discrete parts, such as two or four rectangles, we can approximate the area under the curve.
Each rectangle in this scenario represents a portion of the total area. The width of these sections is determined by dividing the total interval length by the number of rectangles used.
As the number of rectangles increases, the approximation is generally more exact, reducing the gap between the lower and upper sums. This process forms the basis of integral approximation techniques used extensively in calculus.
Rectangular Approximation Method
The Rectangular Approximation Method can be an intuitive way to understand definite integrals. It is performed by using rectangles to estimate the area under a curve such as \( f(x) = \frac{1}{x} \) from \( x = 1 \) to \( x = 5 \).
There are different approaches when deploying this technique:
  • Left Endpoint approximation: Uses the leftmost value in each subinterval to set the rectangle鈥檚 height, often resulting in lower sums.
  • Right Endpoint approximation: Uses the rightmost point, leading to upper sums.
  • Midpoint approximation: Though not used in this initial exercise, it calculates based on the midpoint of each subinterval for potentially more accurate results.
Regardless of method, these approximations help visualize and compute areas that would otherwise be complex to ascertain analytically, and underscoring calculus' practical applications.

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Most popular questions from this chapter

Show that if \(f\) is continuous, then $$\int_{0}^{1} f(x) d x=\int_{0}^{1} f(1-x) d x$$

You will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$f(x)=\frac{x^{4}}{2}-3 x^{3}+10, \quad g(x)=8-12 x$$

Suppose that $$\int_{0}^{1} f(x) d x=3$$ Find $$\int_{-1}^{0} f(x) d x$$ if a. \(f\) is odd, \(\quad\) b. \(f\) is even.

Each of the following functions solves one of the initial value.Which function solves which problem? Give brief reasons for your answers. a. \(y=\int_{1}^{x} \frac{1}{t} d t-3\) b. \(y=\int_{0}^{x} \sec t d t+4\) c. \(y=\int_{-1}^{x} \sec t d t+4\) d. \(y=\int_{\pi}^{x} \frac{1}{t} d t-3\) $$\frac{d y}{d x}=\frac{1}{x}, \quad y(\pi)=-3.$$

Use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100,200,\) and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=\) (average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$f(x)=x \sin ^{2} \frac{1}{x} \quad \text { on } \quad\left[\frac{\pi}{4}, \pi\right]$$
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