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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.1.71c
Chapter 5, Problem 5.1.71c

Displacement from a velocity graph Consider the velocity function for an object moving along a line (see figure).
(c) Use geometry to find the displacement of the object between t = 2 and t = 5.
Velocity graph showing time in seconds on the x-axis and velocity in meters per second on the y-axis.

Verified step by step guidance
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Step 1: Understand the problem. The displacement of the object between t = 2 and t = 5 can be found by calculating the area under the velocity graph between these time intervals. The graph consists of geometric shapes (rectangles and triangles) that can be used to compute the area.
Step 2: Identify the shapes under the graph between t = 2 and t = 5. From t = 2 to t = 3, the graph forms a rectangle. From t = 3 to t = 5, the graph forms a triangle.
Step 3: Calculate the area of the rectangle from t = 2 to t = 3. The height of the rectangle is the velocity at t = 2, which is 20 m/s, and the width is the time interval, which is 1 second. Use the formula for the area of a rectangle: A=l×w.
Step 4: Calculate the area of the triangle from t = 3 to t = 5. The base of the triangle is the time interval, which is 2 seconds, and the height is the change in velocity, which is 20 m/s - 10 m/s = 10 m/s. Use the formula for the area of a triangle: A=b×h2.
Step 5: Add the areas of the rectangle and triangle to find the total displacement of the object between t = 2 and t = 5. The sum of these areas represents the total displacement in meters.

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Key Concepts

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

Velocity Function

The velocity function describes how the velocity of an object changes over time. It is a graphical representation where the y-axis represents velocity (in meters per second) and the x-axis represents time (in seconds). Understanding this function is crucial for analyzing the motion of the object, as it provides insights into how fast and in what direction the object is moving at any given moment.
Recommended video:
10:17
Using The Velocity Function

Displacement

Displacement is the overall change in position of an object over a specific time interval. It can be calculated by finding the area under the velocity graph between two time points. In this case, the displacement from t = 2 to t = 5 seconds can be determined by calculating the area of the geometric shapes formed by the velocity graph within that interval, which represents the total distance traveled in that time frame.
Recommended video:
10:17
Using The Velocity Function

Area Under the Curve

In calculus, the area under a curve on a graph represents the integral of a function over a specified interval. For a velocity graph, this area corresponds to the displacement of the object. By using geometric shapes such as rectangles and triangles, one can calculate the area under the curve between two time points, which provides the total displacement of the object during that time period.
Recommended video:
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Estimating the Area Under a Curve with Right Endpoints & Midpoint
Related Practice
Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(c) The average value of a linear function on an interval [a, b] is the function value at the midpoint of [a, b] .

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Textbook Question

Zero net area Consider the function ƒ(𝓍) = 𝓍² ― 4𝓍 .                                                                                                                                       

                                                                                                                                                                                     c) In general, for the function ƒ(𝓍) = 𝓍² ― a𝓍, where a > 0, for what value of b > 0 (as a function of a) is ∫₀ᵇ ƒ(𝓍) d𝓍 = 0 ? 

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Textbook Question

Matching functions with area functions Match the functions ƒ, whose graphs are given in a― d, with the area functions A (𝓍) = ∫₀ˣ ƒ(t) dt, whose graphs are given in A–D.



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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.                                                                          

                                                                                                                                                                                     (c) The functions p(𝓍) = sin 3𝓍 and q(𝓍) = 4 sin 3𝓍 are antiderivatives of the same function. 

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Textbook Question

Use Table 5.6 to evaluate the following definite integrals.                                                                                                                    

 (c) ∫₃√₂^⁶ d𝓍/(𝓍² ―9)

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Textbook Question

{Use of Tech} Approximating definite integrals Complete the following steps for the given integral and the given value of n. 

(c) Calculate the left and right Riemann sums for the given value of n.

∫₁⁷ 1/𝓍 d𝓍 ; n = 6

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