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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Briggs 3rd EditionCh. 5 - Integration문제 5.2.57a
5장, 문제 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

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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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주요 개념

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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.
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06:21
Properties of Functions

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.
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가이드 코스
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Definition of the Definite Integral

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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가이드 코스
06:18
Integration by Parts for Definite Integrals
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ƒ(t) = 2t + 5 , a = 0

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(a) Find the mass of the left half of the rod (0 ≤ x ≤ 5) .

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 (a) ∫ e¹⁰ˣ d𝓍

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(a) A(2)

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Area functions for the same linear function Let ƒ(t) = 2t ― 2 and consider the two area functions A (𝓍) = ∫₁ˣ ƒ(t) dt and F(𝓍) = ∫₄ˣ ƒ(t) dt .

(a) Evaluate A (2) and A (3). Then use geometry to find an expression for A (𝓍) , for 𝓍 ≥ 1 .

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