To solve the problem of minimizing the integral \( I(a) = \int_0^1 |x^n - a^n| \, dx \) for \( a \in (0, 1) \), we can follow these steps:
### Step 1: Split the Integral
Since \( a \in (0, 1) \), we can split the integral at \( x = a \):
\[
I(a) = \int_0^a |x^n - a^n| \, dx + \int_a^1 |x^n - a^n| \, dx
\]
In the interval \( [0, a] \), \( x^n < a^n \), so \( |x^n - a^n| = a^n - x^n \). In the interval \( [a, 1] \), \( x^n > a^n \), so \( |x^n - a^n| = x^n - a^n \).
Thus, we can rewrite the integral as:
\[
I(a) = \int_0^a (a^n - x^n) \, dx + \int_a^1 (x^n - a^n) \, dx
\]
### Step 2: Evaluate the Integrals
Now we evaluate each integral separately.
1. For the first integral:
\[
\int_0^a (a^n - x^n) \, dx = a^n x \bigg|_0^a - \frac{x^{n+1}}{n+1} \bigg|_0^a = a^{n+1} - \frac{a^{n+1}}{n+1} = a^{n+1} \left(1 - \frac{1}{n+1}\right) = \frac{n a^{n+1}}{n+1}
\]
2. For the second integral:
\[
\int_a^1 (x^n - a^n) \, dx = \frac{x^{n+1}}{n+1} \bigg|_a^1 - a^n (x \bigg|_a^1) = \frac{1}{n+1} - \frac{a^{n+1}}{n+1} - a^n (1 - a) = \frac{1 - a^{n+1}}{n+1} - a^n (1 - a)
\]
### Step 3: Combine the Results
Now we combine the results:
\[
I(a) = \frac{n a^{n+1}}{n+1} + \left(\frac{1 - a^{n+1}}{n+1} - a^n (1 - a)\right)
\]
Simplifying this gives:
\[
I(a) = \frac{n a^{n+1} + 1 - a^{n+1} - (1 - a) a^n}{n+1}
\]
\[
= \frac{(n - 1) a^{n+1} + 1 - a^n + a^{n+1}}{n+1}
\]
\[
= \frac{(n a^{n+1} - a^n + 1)}{n+1}
\]
### Step 4: Differentiate and Find Critical Points
To minimize \( I(a) \), we differentiate with respect to \( a \):
\[
\frac{dI}{da} = \frac{d}{da} \left( \frac{(n a^{n+1} - a^n + 1)}{n+1} \right)
\]
Using the product and chain rules, we find:
\[
\frac{dI}{da} = \frac{n(n+1)a^n - n a^{n-1}}{n+1}
\]
Setting this equal to zero to find critical points:
\[
n(n+1)a^n - n a^{n-1} = 0
\]
Factoring out \( n a^{n-1} \):
\[
n a^{n-1} (a - \frac{1}{2}) = 0
\]
Thus, \( a = \frac{1}{2} \) is a critical point.
### Step 5: Conclusion
The value of \( a \) that minimizes the integral \( I(a) \) is:
\[
\boxed{\frac{1}{2}}
\]
