Sure, let's solve this step by step given the integrals:
[tex]\[
\int_{4}^{1} f(x) \, dx = 15 \quad \text{and} \quad \int_{1}^{3} f(x) \, dx = -14
\][/tex]
First, we need to find [tex]\(\int_{4}^{3} f(x) \, dx\)[/tex].
The integral from 4 to 1 can be split as follows:
[tex]\[
\int_{4}^{1} f(x) \, dx = \int_{4}^{3} f(x) \, dx + \int_{3}^{1} f(x) \, dx
\][/tex]
We'll solve for [tex]\(\int_{4}^{3} f(x) \, dx\)[/tex]. Notice that [tex]\(\int_{3}^{1} f(x) \, dx\)[/tex] is the negative of [tex]\(\int_{1}^{3} f(x) \, dx\)[/tex]:
[tex]\[
\int_{3}^{1} f(x) \, dx = -\int_{1}^{3} f(x) \, dx = -(-14) = 14
\][/tex]
Now we have:
[tex]\[
\int_{4}^{1} f(x) \, dx = \int_{4}^{3} f(x) \, dx + 14
\][/tex]
Given that [tex]\(\int_{4}^{1} f(x) \, dx = 15\)[/tex], substituting into the equation, we get:
[tex]\[
15 = \int_{4}^{3} f(x) \, dx + 14
\][/tex]
Solving for [tex]\(\int_{4}^{3} f(x) \, dx\)[/tex]:
[tex]\[
\int_{4}^{3} f(x) \, dx = 15 - 14 = 1
\][/tex]
[tex]\(\int_{4}^{3} f(x) \, dx\)[/tex] thus equals 1.
Now, we need [tex]\(\int_{3}^{4} f(x) \, dx\)[/tex]. We know that:
[tex]\[
\int_{4}^{3} f(x) \, dx = -\int_{3}^{4} f(x) \, dx
\][/tex]
Therefore,
[tex]\[
\int_{3}^{4} f(x) \, dx = -\int_{4}^{3} f(x) \, dx = -1
\][/tex]
Next, to find the integral [tex]\(\int_{3}^{4} 4 f(x) \, dx\)[/tex], we use the fact that multiplying the function [tex]\( f(x) \)[/tex] by a scalar [tex]\( c \)[/tex] will multiply the result of the integral by that scalar:
[tex]\[
\int_{3}^{4} 4 f(x) \, dx = 4 \cdot \int_{3}^{4} f(x) \, dx
\][/tex]
Substituting the value we found:
[tex]\[
\int_{3}^{4} 4 f(x) \, dx = 4 \cdot (-1) = -4
\][/tex]
So the value of the integral is:
[tex]\[
\int_{3}^{4} 4 f(x) \, dx = -4
\][/tex]
