Question

in problems 16 and 17 let ( f ) be a continuous function on ( (-infty, infty) ) such that ( int_{1}^{3} f(x) dx = -5 ) and ( int_{3}^{7} f(x) dx = 6 ). evaluate the given integrals.

1. ( int_{1}^{7} f(x) dx )
2. ( int_{1}^{7} -5f(x) dx )

Explanation:

Step1: Recall Integral Properties

We know the property of definite integrals: $\int_{a}^{b} f(x)dx = -\int_{b}^{a} f(x)dx$ and $\int_{a}^{b} kf(x)dx = k\int_{a}^{b} f(x)dx$ (where $k$ is a constant). Also, $\int_{a}^{c} f(x)dx+\int_{c}^{b} f(x)dx=\int_{a}^{b} f(x)dx$. First, let's handle problem 16. We are given $\int_{1}^{3}f(x)dx = - 5$ and $\int_{3}^{7}f(x)dx=6$. We need to find $\int_{1}^{7}f(x)dx$.
Using the property $\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx = \int_{a}^{b}f(x)dx$ with $a = 1$, $c=3$, $b = 7$, we have $\int_{1}^{7}f(x)dx=\int_{1}^{3}f(x)dx+\int_{3}^{7}f(x)dx$.
Substitute $\int_{1}^{3}f(x)dx=-5$ and $\int_{3}^{7}f(x)dx = 6$ into the above formula: $\int_{1}^{7}f(x)dx=-5 + 6=1$.

Step2: Solve Problem 17

We need to find $\int_{1}^{7}-5f(x)dx$. Using the constant multiple rule of integrals $\int_{a}^{b}kf(x)dx=k\int_{a}^{b}f(x)dx$ (here $k=-5$, $a = 1$, $b = 7$), we get $\int_{1}^{7}-5f(x)dx=-5\int_{1}^{7}f(x)dx$.
From problem 16, we know that $\int_{1}^{7}f(x)dx = 1$. Substitute this value into the formula: $-5\times1=-5$.

Answer:

Problem 16: $\boldsymbol{1}$; Problem 17: $\boldsymbol{-5}$