Let
\[f(x)=\int_0^1 |t-x|t \, dt\]
for all real $x$. Sketch the graph of $f(x)$. What is the minimum value of $f(x)$?
To solve the problem, we need to analyze the function \( f(x) = \int_0^1 |t - x| t \, dt \) and find its minimum value. We will also sketch the graph of \( f(x) \).

1. **Break the integral into cases based on the value of \( x \):**
   - For \( x \leq 0 \):
     \[
     f(x) = \int_0^1 (x - t)t \, dt = x \int_0^1 t \, dt - \int_0^1 t^2 \, dt = x \left[ \frac{t^2}{2} \right]_0^1 - \left[ \frac{t^3}{3} \right]_0^1 = x \cdot \frac{1}{2} - \frac{1}{3} = \frac{x}{2} - \frac{1}{3}
     \]
   - For \( x \geq 1 \):
     \[
     f(x) = \int_0^1 (t - x)t \, dt = \int_0^1 t^2 \, dt - x \int_0^1 t \, dt = \left[ \frac{t^3}{3} \right]_0^1 - x \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{3} - x \cdot \frac{1}{2} = \frac{1}{3} - \frac{x}{2}
     \]
   - For \( 0 < x < 1 \):
     \[
     f(x) = \int_0^x (x - t)t \, dt + \int_x^1 (t - x)t \, dt
     \]
     Calculate each part separately:
     \[
     \int_0^x (x - t)t \, dt = x \int_0^x t \, dt - \int_0^x t^2 \, dt = x \left[ \frac{t^2}{2} \right]_0^x - \left[ \frac{t^3}{3} \right]_0^x = x \cdot \frac{x^2}{2} - \frac{x^3}{3} = \frac{x^3}{2} - \frac{x^3}{3} = \frac{x^3}{6}
     \]
     \[
     \int_x^1 (t - x)t \, dt = \int_x^1 t^2 \, dt - x \int_x^1 t \, dt = \left[ \frac{t^3}{3} \right]_x^1 - x \left[ \frac{t^2}{2} \right]_x^1 = \left( \frac{1}{3} - \frac{x^3}{3} \right) - x \left( \frac{1}{2} - \frac{x^2}{2} \right) = \frac{1}{3} - \frac{x^3}{3} - \frac{x}{2} + \frac{x^3}{2}
     \]
     Combine the results:
     \[
     f(x) = \frac{x^3}{6} + \left( \frac{1}{3} - \frac{x^3}{3} - \frac{x}{2} + \frac{x^3}{2} \right) = \frac{x^3}{6} + \frac{1}{3} - \frac{x^3}{3} - \frac{x}{2} + \frac{x^3}{2} = \frac{1}{3} - \frac{x}{2} + \frac{x^3}{3}
     \]

2. **Find the derivative \( f'(x) \) and analyze it:**
   - For \( x < 0 \):
     \[
     f'(x) = \frac{d}{dx} \left( \frac{x}{2} - \frac{1}{3} \right) = \frac{1}{2}
     \]
   - For \( x > 1 \):
     \[
     f'(x) = \frac{d}{dx} \left( \frac{1}{3} - \frac{x}{2} \right) = -\frac{1}{2}
     \]
   - For \( 0 < x < 1 \):
     \[
     f'(x) = \frac{d}{dx} \left( \frac{1}{3} - \frac{x}{2} + \frac{x^3}{3} \right) = -\frac{1}{2} + x^2
     \]
     Set \( f'(x) = 0 \) to find critical points:
     \[
     -\frac{1}{2} + x^2 = 0 \implies x^2 = \frac{1}{2} \implies x = \frac{\sqrt{2}}{2}
     \]

3. **Evaluate \( f(x) \) at the critical point \( x = \frac{\sqrt{2}}{2} \):**
   \[
   f\left( \frac{\sqrt{2}}{2} \right) = \frac{1}{3} - \frac{\frac{\sqrt{2}}{2}}{2} + \frac{\left( \frac{\sqrt{2}}{2} \right)^3}{3} = \frac{1}{3} - \frac{\sqrt{2}}{4} + \frac{2\sqrt{2}}{24} = \frac{1}{3} - \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{12} = \frac{1}{3} - \frac{3\sqrt{2}}{12} + \frac{\sqrt{2}}{12} = \frac{1}{3} - \frac{2\sqrt{2}}{12} = \frac{1}{3} - \frac{\sqrt{2}}{6} = \frac{2 - \sqrt{2}}{6}
   \]

4. **Conclusion:**
   The minimum value of \( f(x) \) occurs at \( x = \frac{\sqrt{2}}{2} \) and the minimum value is \( \frac{2 - \sqrt{2}}{6} \).

The final answer is \( \boxed{\frac{2 - \sqrt{2}}{6}} \).