The problem asks to find the first-degree function $f(x)$ that satisfies the following two conditions:

1. $\int_0^1 f(x) \, dx = -1$
2. $\int_0^1 x f(x) \, dx = 0$

Step 1: Assume $f(x)$ is a first-degree (linear) function

Let $f(x) = ax + b$, where $a$ and $b$ are constants to be determined.

Step 2: Apply the first condition

The first condition is: $\int_0^1 f(x) \, dx = -1$ Substitute $f(x) = ax + b$ into the integral: $\int_0^1 (ax + b) \, dx = -1$ This becomes: $\int_0^1 ax \, dx + \int_0^1 b \, dx = -1$ Evaluating these integrals: $\frac{a}{2} + b = -1 \quad \text{(Equation 1)}$

Step 3: Apply the second condition

The second condition is: $\int_0^1 x f(x) \, dx = 0$ Substitute $f(x) = ax + b$: $\int_0^1 x(ax + b) \, dx = 0$ This becomes: $\int_0^1 (ax^2 + bx) \, dx = 0$ Evaluate the integrals: $\frac{a}{3} + \frac{b}{2} = 0 \quad \text{(Equation 2)}$

Step 4: Solve the system of equations

We now solve the system of equations:

1. $\frac{a}{2} + b = -1$
2. $\frac{a}{3} + \frac{b}{2} = 0$

Multiply Equation 1 by 3 and Equation 2 by 2 to eliminate fractions:

1. $3\left(\frac{a}{2} + b\right) = 3(-1) \quad \Rightarrow \quad \frac{3a}{2} + 3b = -3$
2. $2\left(\frac{a}{3} + \frac{b}{2}\right) = 2(0) \quad \Rightarrow \quad \frac{2a}{3} + b = 0$

Now solve this system of equations to find $a$ and $b$.

Let me calculate these values for you.The solution to the system of equations is:

$a = 6 \quad \text{and} \quad b = -4$

Thus, the first-degree function $f(x)$ is:

$f(x) = 6x - 4$

This is the required function that satisfies both conditions.

Would you like more details or have any questions?

Here are 5 related questions:

1. How can we interpret the function $f(x) = 6x - 4$ graphically?
2. What happens if we change the limits of integration in the problem?
3. How do we find higher-degree polynomial solutions for similar problems?
4. Can you explain the geometric meaning of the second condition, $\int_0^1 x f(x) \, dx = 0$?
5. How would the solution change if the conditions involved derivatives?

Tip: Always check if the conditions require higher-degree polynomials or constraints on derivatives before assuming a linear function.