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

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