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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.

次の条件を満たす1次関数 f(x) を求めよ: \( \int_0^1 f(x) \, dx = -1, \int_0^1 x f(x) \, dx = 0 \)

Solve for the first-degree function f(x) that satisfies two integral conditions: \( \int_0^1 f(x) dx = -1 \) and \( \int_0^1 x f(x) dx = 0 \). This problem involves linear functions and basic calculus concepts.

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