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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.2.26b

The following functions are positive and negative on the given interval.
Ζ’(𝓍) = xe⁻ˣ on [-1,1]
(b) Approximate the net area bounded by the graph of f and the x-axis on the interval using a left, right, and midpoint Riemann sum with n = 4.

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1
Step 1: Divide the interval [-1, 1] into n = 4 subintervals. The width of each subinterval, Δ𝓍, is calculated as Δ𝓍 = (1 - (-1)) / 4 = 2 / 4 = 0.5.
Step 2: Identify the endpoints of the subintervals. The subintervals are [-1, -0.5], [-0.5, 0], [0, 0.5], and [0.5, 1].
Step 3: For the left Riemann sum, use the left endpoint of each subinterval to evaluate the function Ζ’(𝓍). The sum is given by: Ξ£ (Δ𝓍 * Ζ’(left endpoint)).
Step 4: For the right Riemann sum, use the right endpoint of each subinterval to evaluate the function Ζ’(𝓍). The sum is given by: Ξ£ (Δ𝓍 * Ζ’(right endpoint)).
Step 5: For the midpoint Riemann sum, use the midpoint of each subinterval to evaluate the function Ζ’(𝓍). The sum is given by: Ξ£ (Δ𝓍 * Ζ’(midpoint)).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Riemann Sums

Riemann sums are a method for approximating the area under a curve by dividing the interval into smaller subintervals. For each subinterval, a sample point is chosen, and the function's value at that point is multiplied by the width of the subinterval. The sum of these products gives an approximation of the total area. There are different types of Riemann sums, including left, right, and midpoint sums, which differ based on the choice of sample points.
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Left, Right, and Midpoint Riemann Sums

In a left Riemann sum, the function's value at the left endpoint of each subinterval is used to calculate the area. Conversely, a right Riemann sum uses the right endpoint. A midpoint Riemann sum takes the function's value at the midpoint of each subinterval. Each method provides a different approximation of the area, and the choice of method can affect the accuracy of the approximation.
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Left, Right, & Midpoint Riemann Sums

Net Area

The net area refers to the total area between the graph of a function and the x-axis over a specified interval, accounting for areas above the x-axis as positive and areas below as negative. In the context of Riemann sums, calculating the net area involves summing the contributions from both positive and negative sections of the function within the interval. This concept is crucial for understanding how the function behaves over the interval and how it affects the overall area calculation.
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Related Practice
Textbook Question

Area functions for the same linear function Let Ζ’(t) = t and consider the two area functions A(𝓍) = βˆ«β‚€Λ£ Ζ’(t) dt and F(𝓍) = βˆ«β‚‚Λ£ Ζ’(t) dt .

(b) Evaluate F(4) and F(6). Then use geometry to find an expression for F (𝓍) , for 𝓍 β‰₯ 2. 

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Textbook Question

Area functions for linear functions Consider the following functions Ζ’ and real numbers a (see figure).


b) Verify that A'(𝓍) = Ζ’(𝓍).



Ζ’(t) = 4t + 2 , a = 0

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Textbook Question

Suppose Ζ’ is an even function and βˆ«βΈβ‚‹β‚ˆ Ζ’(𝓍) d𝓍 = 18

(b) Evaluate βˆ«β‚‹β‚ˆβΈ 𝓍ƒ(𝓍) d𝓍 .

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Textbook Question

Sigma notation Express the following sums using sigma notation. (Answers are not unique.)

(a) 1 + 2 + 3 + 4 + 5

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Textbook Question

Mass from density A thin 10-cm rod is made of an alloy whose density varies along its length according to the function shown in the figure. Assume density is measured in units of g/cm. In Chapter 6, we show that the mass of the rod is the area under the density curve.

(b) Find the mass of the right half of the rod (5 ≀ x ≀ 10) .

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Textbook Question

{Use of Tech} Approximating net area The following functions are positive and negative on the given interval.

f(𝓍) = xΒ³ on [-1,2]

(b) Approximate the net area bounded by the graph of f and the x-axis on the interval using a left, right, and midpoint Riemann sum with n = 4.

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