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Math  /  Calculus

QuestionSuppose ∫03f(x)dx=2,∫35f(x)dx=−4\int_{0}^{3} f(x) d x=2, \int_{3}^{5} f(x) d x=-4, and ∫35g(x)dx=1\int_{3}^{5} g(x) d x=1. Evaluate the integrals in parts a −d-d. a. ∫034f(x)dx=8\int_{0}^{3} 4 f(x) d x=8 (Simplify your answer.) b. ∫35−9g(x)dx=□\int_{3}^{5}-9 g(x) d x=\square (Simplify your answer.)

Studdy Solution

STEP 1

1. We are given definite integrals of functions f(x) f(x) and g(x) g(x) over specific intervals.
2. The problem involves evaluating integrals using properties of definite integrals, such as linearity.

STEP 2

1. Evaluate the integral in part (a) using the linearity property of integrals.
2. Evaluate the integral in part (b) using the linearity property of integrals.

STEP 3

For part (a), use the linearity property of integrals, which states that for a constant c c , ∫abc⋅f(x) dx=c⋅∫abf(x) dx\int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx.
Given: ∫03f(x) dx=2\int_{0}^{3} f(x) \, dx = 2
We need to evaluate: ∫034f(x) dx\int_{0}^{3} 4 f(x) \, dx
Apply the linearity property: ∫034f(x) dx=4⋅∫03f(x) dx=4⋅2=8\int_{0}^{3} 4 f(x) \, dx = 4 \cdot \int_{0}^{3} f(x) \, dx = 4 \cdot 2 = 8
The answer is already simplified as given: 8\boxed{8}

STEP 4

For part (b), use the linearity property of integrals.
Given: ∫35g(x) dx=1\int_{3}^{5} g(x) \, dx = 1
We need to evaluate: ∫35−9g(x) dx\int_{3}^{5} -9 g(x) \, dx
Apply the linearity property: ∫35−9g(x) dx=−9⋅∫35g(x) dx=−9⋅1=−9\int_{3}^{5} -9 g(x) \, dx = -9 \cdot \int_{3}^{5} g(x) \, dx = -9 \cdot 1 = -9
The simplified answer is: −9\boxed{-9}

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