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Math  /  Calculus

QuestionEvaluate each definite integral if it is known that ∫03f(x)dx=−5, and ∫03g(x)dx=3\begin{array}{l} \int_{0}^{3} f(x) d x=-5, \text { and } \\ \int_{0}^{3} g(x) d x=3 \end{array}
Utilize properties of definite integrals to evaluate ∫03[2f(x)+3g(x)]dx\int_{0}^{3}[2 f(x)+3 g(x)] d x

Studdy Solution

STEP 1

1. We are given the values of two definite integrals: ∫03f(x) dx=−5\int_{0}^{3} f(x) \, dx = -5 and ∫03g(x) dx=3\int_{0}^{3} g(x) \, dx = 3.
2. We need to use the linearity property of definite integrals to evaluate ∫03[2f(x)+3g(x)] dx\int_{0}^{3} [2f(x) + 3g(x)] \, dx.

STEP 2

1. Apply the linearity property of integrals.
2. Evaluate the integral using given values.

STEP 3

Apply the linearity property of integrals. The linearity property states that the integral of a sum is the sum of the integrals, and constants can be factored out of the integral:
∫03[2f(x)+3g(x)] dx=∫032f(x) dx+∫033g(x) dx\int_{0}^{3} [2f(x) + 3g(x)] \, dx = \int_{0}^{3} 2f(x) \, dx + \int_{0}^{3} 3g(x) \, dx

STEP 4

Factor out the constants from each integral:
=2∫03f(x) dx+3∫03g(x) dx= 2 \int_{0}^{3} f(x) \, dx + 3 \int_{0}^{3} g(x) \, dx

STEP 5

Substitute the given values of the integrals:
=2(−5)+3(3)= 2(-5) + 3(3)

STEP 6

Perform the arithmetic operations:
=−10+9= -10 + 9 =−1= -1
The value of the definite integral ∫03[2f(x)+3g(x)] dx\int_{0}^{3} [2f(x) + 3g(x)] \, dx is:
−1\boxed{-1}

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