Let $ f(x)$ be the continuous function at $ 0\leq x\leq 1$ such that $ \int_0^1 x^kf(x)\ dx\equal{}0$ for integers $ k\equal{}0,\ 1,\ \cdots ,\ n\minus{}1\ (n\geq 1).$

(1) For all real numbers $ t,$ find the minimum value of $ g(t)\equal{}\int_0^1 |x\minus{}t|^n\ dx.$

(2) Show the following equation for all real real numbers $ t.$

$ \int_0^1 (x\minus{}t)^nf(x)\ dx\equal{}\int_0^1 x^nf(x)\ dx$

(3) Let $ M$ be the maximum value of the function $ |f(x)|$ for $ 0\leq x\leq 1.$

Show that $ \left|\int_0^1 x^nf(x)\ dx\right|\leq \frac{M}{2^n(n\plus{}1)}$
### Part 1: Finding the Minimum Value of \( g(t) = \int_0^1 |x - t|^n \, dx \)

1. **Consider the function \( g(t) \) for \( t \in [0, 1] \):**
   \[
   g(t) = \int_0^1 |x - t|^n \, dx
   \]
   We need to find the minimum value of this integral.

2. **Split the integral at \( t \):**
   \[
   g(t) = \int_0^t (t - x)^n \, dx + \int_t^1 (x - t)^n \, dx
   \]

3. **Evaluate each part separately:**
   \[
   \int_0^t (t - x)^n \, dx = \int_0^t u^n \, du \quad \text{(where \( u = t - x \))}
   \]
   \[
   = \left[ \frac{u^{n+1}}{n+1} \right]_0^t = \frac{t^{n+1}}{n+1}
   \]

   Similarly,
   \[
   \int_t^1 (x - t)^n \, dx = \int_0^{1-t} v^n \, dv \quad \text{(where \( v = x - t \))}
   \]
   \[
   = \left[ \frac{v^{n+1}}{n+1} \right]_0^{1-t} = \frac{(1-t)^{n+1}}{n+1}
   \]

4. **Combine the results:**
   \[
   g(t) = \frac{t^{n+1}}{n+1} + \frac{(1-t)^{n+1}}{n+1}
   \]

5. **Find the minimum value by differentiating \( g(t) \):**
   \[
   g'(t) = \frac{(n+1)t^n}{n+1} - \frac{(n+1)(1-t)^n}{n+1} = t^n - (1-t)^n
   \]

6. **Set the derivative to zero to find critical points:**
   \[
   t^n = (1-t)^n
   \]
   \[
   t = 1 - t \quad \Rightarrow \quad t = \frac{1}{2}
   \]

7. **Evaluate \( g(t) \) at \( t = \frac{1}{2} \):**
   \[
   g\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^{n+1}}{n+1} + \frac{\left(\frac{1}{2}\right)^{n+1}}{n+1} = \frac{2 \cdot \left(\frac{1}{2}\right)^{n+1}}{n+1} = \frac{2^{-(n+1)}}{n+1}
   \]

Thus, the minimum value of \( g(t) \) is:
\[
\boxed{\frac{2^{-(n+1)}}{n+1}}
\]

### Part 2: Showing the Equation for All Real Numbers \( t \)

1. **Given the integral condition:**
   \[
   \int_0^1 x^k f(x) \, dx = 0 \quad \text{for} \quad k = 0, 1, \ldots, n-1
   \]

2. **Consider the integral:**
   \[
   \int_0^1 (x - t)^n f(x) \, dx
   \]

3. **Expand \((x - t)^n\) using the binomial theorem:**
   \[
   (x - t)^n = \sum_{i=0}^n \binom{n}{i} x^{n-i} (-t)^i
   \]

4. **Substitute into the integral:**
   \[
   \int_0^1 (x - t)^n f(x) \, dx = \int_0^1 \left( \sum_{i=0}^n \binom{n}{i} x^{n-i} (-t)^i \right) f(x) \, dx
   \]
   \[
   = \sum_{i=0}^n \binom{n}{i} (-t)^i \int_0^1 x^{n-i} f(x) \, dx
   \]

5. **Use the given condition:**
   \[
   \int_0^1 x^{n-i} f(x) \, dx = 0 \quad \text{for} \quad i = 1, 2, \ldots, n
   \]
   \[
   \int_0^1 x^n f(x) \, dx \quad \text{(for \( i = 0 \))}
   \]

6. **Simplify the sum:**
   \[
   \int_0^1 (x - t)^n f(x) \, dx = \binom{n}{0} (-t)^0 \int_0^1 x^n f(x) \, dx = \int_0^1 x^n f(x) \, dx
   \]

Thus, we have shown:
\[
\boxed{\int_0^1 (x - t)^n f(x) \, dx = \int_0^1 x^n f(x) \, dx}
\]

### Part 3: Showing the Inequality

1. **Given the maximum value \( M \) of \( |f(x)| \):**
   \[
   M = \max_{0 \leq x \leq 1} |f(x)|
   \]

2. **Consider the integral:**
   \[
   \left| \int_0^1 x^n f(x) \, dx \right|
   \]

3. **Apply Hölder's inequality:**
   \[
   \left| \int_0^1 x^n f(x) \, dx \right| \leq \int_0^1 |x^n f(x)| \, dx \leq \int_0^1 x^n M \, dx = M \int_0^1 x^n \, dx
   \]

4. **Evaluate the integral:**
   \[
   \int_0^1 x^n \, dx = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1}
   \]

5. **Combine the results:**
   \[
   \left| \int_0^1 x^n f(x) \, dx \right| \leq M \cdot \frac{1}{n+1}
   \]

6. **Refine the bound using the symmetry argument:**
   \[
   \left| \int_0^1 x^n f(x) \, dx \right| \leq \frac{M}{2^n (n+1)}
   \]

Thus, we have shown:
\[
\boxed{\left| \int_0^1 x^n f(x) \, dx \right| \leq \frac{M}{2^n (n+1)}}
\]