question let f be a continuous function such that ∫−28f(x)dx = 6 and ∫−25f(x)dx = 1. what is the value of…

Answer:

$-15$

Explanation:

Step1: Use integral property

We know that $\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx$. So, $\int_{-2}^{8}f(x)dx=\int_{-2}^{5}f(x)dx+\int_{5}^{8}f(x)dx$. Given $\int_{-2}^{8}f(x)dx = 6$ and $\int_{-2}^{5}f(x)dx=1$, we can find $\int_{5}^{8}f(x)dx$ as follows: $6 = 1+\int_{5}^{8}f(x)dx$, then $\int_{5}^{8}f(x)dx=6 - 1=5$.

Step2: Use constant - multiple rule of integrals

The constant - multiple rule of integrals states that $\int_{a}^{b}kf(x)dx=k\int_{a}^{b}f(x)dx$ for a constant $k$. We want to find $\int_{8}^{5}3f(x)dx$. First, note that $\int_{8}^{5}3f(x)dx=- \int_{5}^{8}3f(x)dx$ (since $\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$). By the constant - multiple rule, $\int_{5}^{8}3f(x)dx = 3\int_{5}^{8}f(x)dx$. Since $\int_{5}^{8}f(x)dx = 5$, then $\int_{5}^{8}3f(x)dx=3\times5 = 15$. So, $\int_{8}^{5}3f(x)dx=-15$.