$$ I = \int _ { 0 } ^ { 1 } x f ^ { \prime } ( x ) d x $$

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To evaluate the integral

$$
I = \int_{0}^{1} x f^{\prime}(x) \, dx,
$$

we can use **integration by parts**. Integration by parts is based on the formula:

$$
\int u \, dv = uv - \int v \, du,
$$

where $$ u $$ and $$ dv $$ are continuously differentiable functions on the interval of integration.

### Step 1: Choose $$ u $$ and $$ dv $$

Let's assign:

$$
\begin{cases}
u = x & \Rightarrow \frac{du}{dx} = 1 \, \Rightarrow \, du = dx, \
dv = f^{\prime}(x) \, dx & \Rightarrow v = f(x).
\end{cases}
$$

### Step 2: Apply Integration by Parts

Substituting into the integration by parts formula:

$$
I = \int_{0}^{1} x f^{\prime}(x) \, dx = \left. x f(x) \right|_{0}^{1} - \int_{0}^{1} f(x) \cdot 1 \, dx.
$$

### Step 3: Evaluate the Boundary Terms

Compute the boundary term $$ \left. x f(x) \right|_{0}^{1} $$:

$$
\left. x f(x) \right|_{0}^{1} = (1 \cdot f(1)) - (0 \cdot f(0)) = f(1) - 0 = f(1).
$$

### Step 4: Combine the Results

Putting it all together:

$$
I = f(1) - \int_{0}^{1} f(x) \, dx.
$$

### Final Result

Thus, the integral simplifies to:

$$
\int_{0}^{1} x f^{\prime}(x) \, dx = f(1) - \int_{0}^{1} f(x) \, dx.
$$

This result expresses the original integral in terms of the function $$ f $$ evaluated at the upper limit and the integral of $$ f $$ itself over the interval $$[0, 1]$$.
The integral simplifies to $$ f(1) - \int_{0}^{1} f(x) \, dx $$.

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