Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.2.57a

Chapter 5, Problem 5.2.57a

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(a) ∫₀¹ (4𝓍―2𝓍³) d𝓍

Verified step by step guidance

Step 1: Recognize that the given integral in part (a), ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍, can be split into two separate integrals using the linearity property of integrals: ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍 = ∫₀¹ 4𝓍 d𝓍 ― ∫₀¹ 2𝓍³ d𝓍.

Step 2: Factor out constants from each integral. For the first term, ∫₀¹ 4𝓍 d𝓍 becomes 4∫₀¹ 𝓍 d𝓍. For the second term, ∫₀¹ 2𝓍³ d𝓍 becomes 2∫₀¹ 𝓍³ d𝓍.

Step 3: Use the given value of I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4 to extract the value of ∫₀¹ 𝓍³ d𝓍. Rewrite I as ∫₀¹ 𝓍³ d𝓍 ― ∫₀¹ 2𝓍 d𝓍 = ―3/4. This equation can be solved to find ∫₀¹ 𝓍³ d𝓍.

Step 4: Compute ∫₀¹ 𝓍 d𝓍 using the power rule for integration. The integral of 𝓍 with respect to 𝓍 is (𝓍²)/2. Evaluate this from 0 to 1 to find the value of ∫₀¹ 𝓍 d𝓍.

Step 5: Substitute the values of ∫₀¹ 𝓍³ d𝓍 and ∫₀¹ 𝓍 d𝓍 into the expression for ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍 = 4∫₀¹ 𝓍 d𝓍 ― 2∫₀¹ 𝓍³ d𝓍 to compute the final result.

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Key Concepts

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

Properties of Integrals

The properties of integrals, such as linearity and the ability to split integrals, are fundamental in calculus. Linearity allows us to factor constants out of integrals and combine integrals of the same limits. This means that if we have an integral of a sum, we can separate it into the sum of integrals, which simplifies calculations.

Definite Integrals

Definite integrals represent the signed area under a curve between two points on the x-axis. The notation ∫ₐᵇ f(x) dx indicates the integral of the function f(x) from a to b. The value of a definite integral can be interpreted as the accumulation of quantities, which is essential for evaluating integrals over specific intervals.

Integration Techniques

Various techniques exist for evaluating integrals, including substitution and integration by parts. In this context, recognizing patterns in the integrand can help simplify the integral. For example, if an integral can be expressed in terms of known integrals, such as the one provided (I), it can be evaluated more easily by leveraging previously calculated values.

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Related Practice

Textbook Question

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

(a) Find and graph the area function A (𝓍) = ∫ₐˣ ƒ(t) dt .

ƒ(t) = 2t + 5 , a = 0

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Textbook Question

Working with area functions Consider the function ƒ and the points a, b, and c.

(a) Find the area function A (𝓍) = ∫ₐˣ ƒ(t) dt using the Fundamental Theorem.

ƒ(𝓍) = ― 12𝓍 (𝓍―1) (𝓍― 2) ; a = 0 , b = 1 , c = 2

Textbook Question

Use Table 5.6 to evaluate the following indefinite integrals.

(a) ∫ e¹⁰ˣ d𝓍

Textbook Question

Area functions The graph of ƒ is shown in the figure. Let A(x) = ∫₋₂ˣ ƒ(t) dt and F(x) = ∫₄ˣ ƒ(t) dt be two area functions for ƒ. Evaluate the following area functions.

(a) A (―2)

Textbook Question

{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.

(a) Write the midpoint Riemann sum in sigma notation for an arbitrary value of n.

∫₁⁴ 2√𝓍 d𝓍

Textbook Question

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

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

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4(a) ∫₀¹ (4𝓍―2𝓍³) d𝓍

Verified video answer for a similar problem:

Key Concepts

Properties of Integrals

Definite Integrals

Integration Techniques

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(a) ∫₀¹ (4𝓍―2𝓍³) d𝓍