Properties of integrals Suppose ∫₀³ƒ(𝓍) d𝓍 = 2 , ∫₃⁶ƒ(𝓍) d𝓍 = ―5 , and ∫₃⁶g(𝓍) d𝓍 = 1. Evaluate the following integrals.
(d) ∫₆³ (ƒ(𝓍) + 2g(𝓍)) d𝓍

검증된 단계별 안내

Step 1: Recognize that the integral ∫₆³ (ƒ(𝓍) + 2g(𝓍)) d𝓍 involves reversing the limits of integration. When the limits are reversed, the integral changes sign. Thus, ∫₆³ (ƒ(𝓍) + 2g(𝓍)) d𝓍 = -∫₃⁶ (ƒ(𝓍) + 2g(𝓍)) d𝓍.

Step 2: Use the property of linearity of integrals to split the integral into two separate integrals: -∫₃⁶ (ƒ(𝓍) + 2g(𝓍)) d𝓍 = -[∫₃⁶ ƒ(𝓍) d𝓍 + ∫₃⁶ 2g(𝓍) d𝓍].

Step 3: Factor out the constant 2 from the second integral using the constant multiple rule: -[∫₃⁶ ƒ(𝓍) d𝓍 + 2∫₃⁶ g(𝓍) d𝓍].

Step 4: Substitute the given values for the integrals: ∫₃⁶ ƒ(𝓍) d𝓍 = -5 and ∫₃⁶ g(𝓍) d𝓍 = 1. Replace these values into the expression: -[-5 + 2(1)].

Step 5: Simplify the expression inside the brackets and apply the negative sign outside the brackets to find the result.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Properties of Definite Integrals

Definite integrals have several key properties, including linearity, which states that the integral of a sum is the sum of the integrals. This means that ∫(f(x) + g(x)) dx = ∫f(x) dx + ∫g(x) dx. Additionally, the integral from a to b can be expressed as the negative of the integral from b to a, i.e., ∫_a^b f(x) dx = -∫_b^a f(x) dx.

Change of Limits in Integrals

When evaluating integrals, changing the limits of integration affects the sign of the result. Specifically, if you reverse the limits of integration, the value of the integral becomes negative. For example, ∫_a^b f(x) dx = -∫_b^a f(x) dx, which is crucial when evaluating integrals with limits that are not in increasing order.

Linear Combination of Functions

A linear combination of functions involves adding or subtracting functions multiplied by constants. In the context of integrals, if you have a function like (f(x) + 2g(x)), you can evaluate the integral of this combination by integrating each function separately and applying the constants accordingly. This property simplifies the evaluation of integrals involving multiple functions.

Properties of integrals Suppose ∫₀³ƒ(𝓍) d𝓍 = 2 , ∫₃⁶ƒ(𝓍) d𝓍 = ―5 , and ∫₃⁶g(𝓍) d𝓍 = 1. Evaluate the following integrals.(d) ∫₆³ (ƒ(𝓍) + 2g(𝓍)) d𝓍