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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.86

Evaluating integrals Evaluate the following integrals.


∫₀⁵ |2𝓍―8|d𝓍

Verified step by step guidance
1
Step 1: Recognize that the integral involves the absolute value function |2𝓍 - 8|. Absolute value functions can be split into piecewise functions based on where the expression inside the absolute value changes sign.
Step 2: Solve for the point where 2𝓍 - 8 = 0. This occurs when 𝓍 = 4. Therefore, the integral can be split into two intervals: [0, 4] and [4, 5].
Step 3: On the interval [0, 4], the expression 2𝓍 - 8 is negative, so |2𝓍 - 8| = -(2𝓍 - 8). On the interval [4, 5], the expression 2𝓍 - 8 is positive, so |2𝓍 - 8| = 2𝓍 - 8.
Step 4: Rewrite the integral as the sum of two integrals: ∫₀⁴ -(2𝓍 - 8)d𝓍 + ∫₄⁵ (2𝓍 - 8)d𝓍. Simplify the integrands in each interval.
Step 5: Compute each integral separately by applying the power rule for integration and evaluating the definite integrals. Combine the results to find the total value of the original integral.

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Key Concepts

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

Definite Integrals

A definite integral calculates the accumulation of a quantity, represented as the area under a curve between two specified limits. In this case, the integral ∫₀⁵ |2𝓍―8|d𝓍 evaluates the area between the curve of the function |2𝓍―8| and the x-axis from x=0 to x=5. Understanding how to set up and evaluate definite integrals is crucial for solving problems involving areas and accumulated quantities.
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Definition of the Definite Integral

Absolute Value Function

The absolute value function, denoted as |x|, outputs the non-negative value of x regardless of its sign. In the integral ∫₀⁵ |2𝓍―8|d𝓍, the expression inside the absolute value, 2𝓍―8, can be positive or negative depending on the value of x. Identifying where the expression changes sign is essential for correctly evaluating the integral, as it affects the area calculation.
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Piecewise Functions

Piecewise functions are defined by different expressions based on the input value. For the integral involving |2𝓍―8|, we need to determine the points where 2𝓍―8 equals zero to split the integral into segments where the function behaves differently. This approach allows us to evaluate the integral accurately by considering the appropriate expression for each segment of the interval.
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Related Practice
Textbook Question

Area Find (i) the net area and (ii) the area of the following regions. Graph the function and indicate the region in question.


The region bounded by y = 6 cos 𝓍 and the 𝓍-axis between 𝓍 = ―π/2 and 𝓍 = π

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Textbook Question

Evaluating integrals Evaluate the following integrals.


∫₁⁴ ((√v + v) / v ) dv

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Textbook Question

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.

(f) Find a constant C such that F(𝓍) = G(𝓍) + C .

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Textbook Question

Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result.


 ∫₀⁴ ƒ(𝓍) d𝓍, where ƒ(𝓍) = {5      if 𝓍 ≤ 2                                                                                                                                                                                     

                      3𝓍 ― 1  if 𝓍 > 2

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Textbook Question

Integration by Riemann sums Consider the integral ∫₁⁴ (3𝓍― 2) d𝓍.


(a) Evaluate the right Riemann sum for the integral with n = 3 .

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Textbook Question

Variations on the substitution method Evaluate the following integrals.                                                                                                        

                                                                                                                                                                    

 ∫ (eˣ ― e⁻ˣ)/ (eˣ + e⁻ˣ) d𝓍

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