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Let's analyze the question presented in the image:

### Problem Summary:
We are given that \( u(x) \) and \( v(x) \) are differentiable functions of \( x \). We are provided with:
\[
\int_0^1 v \, du = 4
\]
along with a table of functional values for \( u(x) \) and \( v(x) \):
\[
\begin{array}{|c|c|c|}
\hline
x & u(x) & v(x) \\
\hline
0 & 2 & -4 \\
1 & 3 & 0 \\
\hline
\end{array}
\]
We are also given that \( \left. \frac{1}{2}v^2 \right|_0^1 = 1 \). The goal is to find \( \int_0^1 u \, dv \).

### Solution Strategy:

We are given the integral \( \int_0^1 v \, du = 4 \) and asked to determine \( \int_0^1 u \, dv \).

#### Using Integration by Parts:
To solve this, we'll apply **integration by parts**, which is given by the formula:
\[
\int u \, dv = uv - \int v \, du
\]
Using this formula, we can rewrite the integral \( \int_0^1 u \, dv \) as:
\[
\int_0^1 u \, dv = \left[ uv \right]_0^1 - \int_0^1 v \, du
\]
Now, substituting the given values:
1. From the table, \( u(1) = 3 \) and \( v(1) = 0 \), so \( u(1)v(1) = 3 \times 0 = 0 \).
2. Also, \( u(0) = 2 \) and \( v(0) = -4 \), so \( u(0)v(0) = 2 \times (-4) = -8 \).

Thus,
\[
\left[ uv \right]_0^1 = 0 - (-8) = 8
\]

We are also given that \( \int_0^1 v \, du = 4 \). Substituting this into the equation for integration by parts:
\[
\int_0^1 u \, dv = 8 - 4 = 4
\]

### Final Answer:
\[
\int_0^1 u \, dv = 4
\]

Would you like further details or explanations on any part of the solution?

### Related Questions:
1. How is integration by parts derived from the product rule of differentiation?
2. Can integration by parts be applied to functions that are not continuously differentiable?
3. What other techniques are useful when evaluating integrals involving multiple functions?
4. How does the relationship between \( u(x) \) and \( v(x) \) affect their integrals?
5. What happens if the limits of integration were different in this case?

### Tip:
Always remember to carefully check boundary values and ensure that all given conditions (such as function values) are used correctly when applying integration techniques like integration by parts.

Suppose that u and v are differentiable functions of x with ∫ v du = 4 and the following functional values: u(0) = 2, u(1) = 3, v(0) = -4, v(1) = 0. Use this information to determine ∫ u dv.

Solve the integral ∫ u dv using the given functional values and the integration by parts technique. This problem demonstrates the application of a key calculus concept with specific functional values of u and v.

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