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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.R.23c

Area by geometry Use geometry to evaluate the following definite integrals, where the graph of ƒ is given in the figure.
(c) ∫₅⁷ ƒ(𝓍) d𝓍
Graph of a piecewise function showing y = f(x) with defined segments, plotted on a grid with x and y axes.

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Observe the graph of the function ƒ(x) over the interval [5, 7]. The graph consists of two linear segments: one from (5, 3) to (6, 0) and another from (6, 0) to (7, -3).
The definite integral ∫₅⁷ ƒ(x) dx represents the net area between the graph of ƒ(x) and the x-axis over the interval [5, 7]. Positive areas are above the x-axis, and negative areas are below the x-axis.
Break the integral into two parts based on the segments of the graph: ∫₅⁶ ƒ(x) dx for the segment from (5, 3) to (6, 0), and ∫₆⁷ ƒ(x) dx for the segment from (6, 0) to (7, -3).
For the segment from (5, 3) to (6, 0), calculate the area of the triangle formed. The base is 1 unit (from x = 5 to x = 6), and the height is 3 units. Use the formula for the area of a triangle: Area = (1/2) × base × height.
For the segment from (6, 0) to (7, -3), calculate the area of the triangle formed. The base is 1 unit (from x = 6 to x = 7), and the height is -3 units (negative because it is below the x-axis). Use the formula for the area of a triangle: Area = (1/2) × base × height. Add the two areas, keeping in mind their signs, to find the total value of the integral.

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Key Concepts

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

Definite Integral

A definite integral represents the signed area under a curve between two points on the x-axis. It is calculated using the limits of integration, which define the interval over which the area is measured. In this context, the definite integral ∫₅⁷ ƒ(𝓍) d𝓍 calculates the area between the graph of the function f(x) and the x-axis from x = 5 to x = 7.
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Definition of the Definite Integral

Piecewise Function

A piecewise function is defined by different expressions based on the input value. In the given graph, f(x) consists of segments that change at specific x-values, creating distinct linear sections. Understanding how to interpret these segments is crucial for accurately calculating the area under the curve for the definite integral.
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Piecewise Functions

Area Under the Curve

The area under the curve in a graph of a function can be interpreted as the integral of that function. For piecewise functions, this area can be computed by breaking it into simpler geometric shapes, such as rectangles and triangles, and summing their areas. This method simplifies the evaluation of the definite integral by leveraging geometric properties.
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Estimating the Area Under a Curve with Right Endpoints & Midpoint
Related Practice
Textbook Question

Displacement from velocity A particle moves along a line with a velocity given by v(t) = 5 sin πt, starting with an initial position s(0) = 0 . Find the displacement of the particle between t = 0 and t = 2 , which is given by s(t) = ∫₀² v(t) dt . Find the distance traveled by the particle during this interval, which is ∫₀² |v(t)| dt .

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Textbook Question

Function defined by an integral Let H (𝓍) = ∫₀ˣ √(4 ― t²) dt, for ― 2 ≤ 𝓍 ≤ 2.

(c) Evaluate H '(2) .

Textbook Question

Area functions and the Fundamental Theorem Consider the function

ƒ(t) = { t      if  ―2 ≤ t < 0

t²/2    if    0 ≤ t ≤ 2

and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

(c) Use the Fundamental Theorem to find an expression for F '(𝓍) for 0 ≤ 𝓍 < 2.

Textbook Question

Evaluating integrals Evaluate the following integrals.


∫₋π/₂^π/² (cos 2𝓍 + cos 𝓍 sin 𝓍 ― 3 sin 𝓍⁵) d𝓍

Textbook Question

Velocity to displacement An object travels on the 𝓍-axis with a velocity given by v(t) = 2t + 5, for 0 ≤ t ≤ 4.


(c) True or false: The object would travel as far as in part (a) if it traveled at its average velocity (a constant), for 0 ≤ t ≤ 4. .

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ and ƒ' are continuous functions for all real numbers.

(b) Given an area function A(𝓍) = ∫ₐˣ ƒ(t) dt and an antiderivative F of ƒ, it follows that A'(𝓍) = F(𝓍) .

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