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

Verified step by step guidance

1. Step 1: Recognize the property of integrals that allows constants to be factored out. Specifically, for any constant c and function f(x), ∫ₐᵇ cƒ(𝓍) d𝓍 = c ∫ₐᵇ ƒ(𝓍) d𝓍.
2. Step 2: Apply this property to the given integral ∫₀³ 5ƒ(𝓍) d𝓍. Here, the constant 5 can be factored out, resulting in 5 ∫₀³ ƒ(𝓍) d𝓍.
3. Step 3: Substitute the value of ∫₀³ ƒ(𝓍) d𝓍 provided in the problem, which is 2.
4. Step 4: Multiply the constant 5 by the value of the integral ∫₀³ ƒ(𝓍) d𝓍 (which is 2) to complete the evaluation.
5. Step 5: The result of the integral ∫₀³ 5ƒ(𝓍) d𝓍 is obtained by performing the multiplication in Step 4.

Key Concepts

Properties of Integrals

The properties of integrals, particularly the linearity property, state that the integral of a constant multiplied by a function can be factored out. This means that ∫a^b kƒ(𝓍) d𝓍 = k∫a^b ƒ(𝓍) d𝓍, where k is a constant. This property simplifies the evaluation of integrals by allowing constants to be taken outside the integral.

Definite Integrals

Definite integrals represent the signed area under a curve between two limits. The notation ∫a^b ƒ(𝓍) d𝓍 indicates the integral of the function ƒ(𝓍) from the lower limit a to the upper limit b. The result of a definite integral is a number that quantifies this area, which can be positive, negative, or zero depending on the function's behavior over the interval.

Additivity of Integrals

The additivity property of integrals states that the integral over an interval can be split into the sum of integrals over subintervals. Specifically, ∫a^c ƒ(𝓍) d𝓍 = ∫a^b ƒ(𝓍) d𝓍 + ∫b^c ƒ(𝓍) d𝓍 for any point b between a and c. This property is useful for evaluating integrals over larger intervals by breaking them down into smaller, manageable parts.